1.कक्षा 12वीं अनिश्चित समाकल (Indefinite Integrals in Class 12th),अनिश्चित समाकलन कक्षा 12 (Indefinite Integral Class 12):

Indefinite Integrals in Class 12th

कक्षा 12वीं अनिश्चित समाकल (Indefinite Integrals in Class 12th) के इस आर्टिकल में अनिश्चित समाकलनों पर आधारित कुछ विशिष्ट सवालों को हल करके समझने का प्रयास करेंगे।
आपको यह जानकारी रोचक व ज्ञानवर्धक लगे तो अपने मित्रों के साथ इस गणित के आर्टिकल को शेयर करें।यदि आप इस वेबसाइट पर पहली बार आए हैं तो वेबसाइट को फॉलो करें और ईमेल सब्सक्रिप्शन को भी फॉलो करें।जिससे नए आर्टिकल का नोटिफिकेशन आपको मिल सके।यदि आर्टिकल पसन्द आए तो अपने मित्रों के साथ शेयर और लाईक करें जिससे वे भी लाभ उठाए।आपकी कोई समस्या हो या कोई सुझाव देना चाहते हैं तो कमेंट करके बताएं।इस आर्टिकल को पूरा पढ़ें।

Also Read This Article:- Some Properties of Definite Integrals

2.कक्षा 12वीं अनिश्चित समाकल के उदाहरण (Indefinite Integrals in Class 12th Illustrations):

1 से 24 तक के प्रश्नों के फलनों का समाकलन कीजिए।
Illustration:1. \frac{1}{x-x^3}
Solution: \frac{1}{x-x^2} \\ I=\int \frac{1}{x-x^3} d x \\ \Rightarrow I=\int \frac{1}{x(1-x)(1+x)} d x
आंशिक भिन्नों में वियोजित करने पर:

\frac{1}{x(1-x)(1+x)}=\frac{A}{x}+\frac{B}{1-x}+\frac{C}{1+x} \\ \Rightarrow \frac{1}{x(1-x)(1+x)} =\frac{\left.A\left(1-x^2\right)+B x(1+x)+C x(1-x\right)}{x(1-x)(1+x)} \\ \Rightarrow 1=A\left(1-x^2\right) +B x(1+x)+C x(1-x)
Put x=1 \\ 1=B(1)(1+1) \Rightarrow B=\frac{1}{2}
put x=0 \\1=A(1-0) \Rightarrow A=1
put x=-1 \\ 1=A\left(1-(-1)^2\right)+C(-1)(1-(-1)) \\ \Rightarrow 1=A(1-1)-C(1+1) \\ \Rightarrow c=-\frac{1}{2} \\ I=\int\left[\frac{1}{x}+\frac{1}{2(1-x)}-\frac{1}{2(1+x)}\right] d x \\ =\int \frac{1}{x} d x+\frac{1}{2} \int \frac{1}{1-x} d x-\frac{1}{2} \int \frac{1}{1+x} d x \\ =\log |x|-\frac{1}{2} \log \left|1-x\right| -\frac{1}{2} \log \left|1+x\right|+c \\ =\frac{1}{2} \times 2 \log |x|-\frac{1}{2} \log \left|1-x^2\right|+c \\ \Rightarrow I=\frac{1}{2} \log \left|x^2\right|-\frac{1}{2} \log \left|1-x^2\right|+C \\ \Rightarrow I=\frac{1}{2} \log \left|\frac{x^2}{1-x^2}\right|+C
Illustration:2. \frac{1}{\sqrt{x+a}+\sqrt{x+b}}
Solution: \frac{1}{\sqrt{x+a}+\sqrt{x+b}} \\ I=\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} d x \\ =\int \frac{\sqrt{x+a}-\sqrt{x+b}}{(\sqrt{x+a}+\sqrt{x+b})(\sqrt{x+a}-\sqrt{x+b})} dx \\ =\int \frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)} d x \\ =\int \frac{\sqrt{x+a}-\sqrt{x+b}}{a-b} d x \\ =\frac{1}{(a-b)}\left[\int \sqrt{x+a} d x-\int \sqrt{x+b} d x\right]+c \\ =\frac{1}{(a-b)}\left[\frac{(x+a)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(x+b)^{\frac{3}{2}}}{\frac{3}{2}}\right]+c \\ \Rightarrow I =\frac{2}{3(a-b)}\left[(x+a)^{\frac{3}{2}}-(x+b)^{\frac{3}{2}}\right]+c
Illustration:3. \frac{1}{x \sqrt{a x-x^2}}
Solution: \frac{1}{x \sqrt{a x-x^2}} \\ I=\int \frac{1}{x \sqrt{a x-x^2}} d x
Put x=\frac{a}{t} \Rightarrow d x=-\frac{a}{t^2} dt \\ \Rightarrow I=\int \frac{1}{\frac{a}{t} \sqrt{a \cdot \frac{a}{t}-\frac{a^2}{t^2}}}\left(-\frac{a}{t^2}\right) dt \\ =-\int \frac{t}{a \sqrt{\frac{1}{t}-\frac{1}{t^2}}} \cdot \frac{1}{t^2} d t \\ =-\frac{1}{a} \int \frac{1}{\frac{\sqrt{t-1}}{t}} \times \frac{1}{t} d t \\ =-\frac{1}{a} \int \frac{1}{\sqrt{t-1}} d t \\ =-\frac{1}{a} \frac{(t-1)^{\frac{1}{2}}}{\frac{1}{2}}+C \\ =-\frac{2}{a}(t-1)^{\frac{1}{2}} +C \\ \Rightarrow I=-\frac{2}{a}\left(\frac{a}{x}-1\right)^{\frac{1}{2}}+C \\ \Rightarrow I=-\frac{2}{a}\left(\frac{a-x}{x}\right)^{\frac{1}{2}}+C
Illustration:4. \frac{1}{x^2\left(x^4+1\right)^{\frac{3}{4}}}
Solution: \frac{1}{x^2\left(x^4+1\right)^{\frac{3}{4}}} \\ I=\int \frac{1}{x^2\left(x^4+1\right)^{\frac{3}{4}}} d x \\ =\int \frac{1}{x^5\left(\frac{1}{x^4}+1\right)^{\frac{3}{4}}} d x
Put \frac{1}{x^4}=t \Rightarrow d t=\frac{-4}{x^5} dx \\ =\int \frac{1}{(t+1)^{\frac{3}{4}}}\left(-\frac{d t}{4}\right) \\ =-\frac{1}{4} \frac{(t+1)^{-\frac{3}{4}+1}}{-\frac{3}{4}+1}+c \\ =-\frac{1}{4} \times \frac{4}{1}(t+1)^{\frac{1}{4}}+C \\ I=-\left(\frac{1}{x^4}+1\right)^{\frac{1}{4}}+C
Illustration:5. \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}
Solution: \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}} \\ I=\int \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}} dx
put x=t^6 \Rightarrow dx=6 t^5 dt \\ I =\int \frac{1}{\left(t^6\right)^{\frac{1}{2}} +\left(t^6\right)^{\frac{1}{3}}}\left(6 t^5 d t\right) \\ =6 \int \frac{t^5}{t^3+t^2} d t \\ =6 \int \frac{t^5}{t^2(t+1)} d t \\ =6 \int \frac{t^3}{t+1} \\ \begin{array}{c|c} & t^2-t+1 \\ \hline t+1 & t^3 \\ & t^3+t^2 \\ & - \quad - \\ \hline & -t^2 \\ & -t^2-t \\ & + \quad + \\ \hline & t \\ & t+1 \\ & - \quad - \\ \hline & -1\end{array} \\ \Rightarrow I=6 \int\left(t^2-t+1-\frac{1}{t+1}\right) d t \\ =6 \int t^2 d t-6 \int t d t+6 \int 1 \cdot d t-6 \int \frac{1}{t+1} d t\\ =6 \cdot \frac{t^3}{3}-6 \cdot \frac{t^2}{2}+6 t-6 \log |1+t|+C \\ \Rightarrow I =2 x^{\frac{1}{2}}-3 x^{\frac{1}{3}}+6 x^{\frac{1}{6}}-6 \log |1+x^{\frac{1}{6}}|+C
Illustration:6. \frac{5 x}{(x+1)\left(x^2+9\right)}
Solution: \frac{5 x}{(x+1)\left(x^2+9\right)} \\ I=\int \frac{5 x}{(x+1)\left(x^2+9\right)} dx
आंशिक भिन्नों में वियोजित करने पर:

\frac{5 x}{(x+1)\left(x^2+9\right)}=\frac{A}{x+1}+\frac{B x+C}{x^2+9} \\ \Rightarrow \frac{5 x}{(x+1)\left(x^2+9\right)} =\frac{A\left(x^2+9\right)+(B x+C)(x+1)}{(x+1)\left(x^2+9\right)} \\ \Rightarrow 5 x=A\left(x^2+9\right)+(B x+C)(x+1)
Put x=-1 \\ \Rightarrow 5(-1)=A\left((-1)^2+9\right) \\ \Rightarrow-5=A(1+9) \Rightarrow A=-\frac{5}{10} =-\frac{1}{2}
Put x=0 \\ 5(0)=A(9)+C(0+1) \\ \Rightarrow 0=-\frac{1}{2} \times 9+C \Rightarrow C=\frac{9}{2}
Put x=1 \\ \Rightarrow 5(1)=A\left(1^2+9\right)+(B \times 1+C)(1+1) \\ \Rightarrow 5=-\frac{1}{2}(1+9)+\left(B+\frac{9}{2}\right) \times 2 \\ \Rightarrow 5=-5+2 B+9 \\ \Rightarrow 2B=10-9 \Rightarrow B=\frac{1}{2} \\ I=\int\left(-\frac{1}{2(x+1)}+\frac{\frac{1}{2} x+\frac{9}{2}}{x^2+9}\right) d x \\ =-\frac{1}{2} \int \frac{1}{x+1} d x+\frac{1}{2} \int \frac{x}{x^2+9} d x+\frac{9}{2} \int \frac{1}{x^2+9} d x \\ =-\frac{1}{2} \log |1+x|+\frac{1}{4} \log \left|x^2+9\right|+\frac{9}{2} \cdot \frac{1}{3} \tan^{-1} \left(\frac{x}{3}\right)+c \\ \Rightarrow I=-\frac{1}{2} \log |1+x|+\frac{1}{4} \log \mid x^2+9\mid+\frac{3}{2} \tan^{-1} \left(\frac{x}{3}\right)+c
Illustration:7. \frac{\sin x}{\sin (x-a)}
Solution: \frac{\sin x}{\sin (x-a)} \\ I=\int \frac{\sin x}{\sin (x-a)} d x
put x-a=t \Rightarrow d x=d t \\ I=\int \frac{\sin (t+a)}{\sin t} d t \\ =\int \frac{\sin t \cos a+\cos t \sin a}{\sin t} d t \\ =\int \frac{\sin t \cos a}{\sin t} d t+\int \frac{\cos t \sin a}{\sin t} d t \\ =\int \cos a d t+\int \cot t \sin a d t \\ =t \cos a+\sin a \log |\sin t|+c_1 \\ \Rightarrow I=(x-a) \cos a+\sin a\log | \sin (x-a) |+c_1 \\ \Rightarrow I=x \cos a+\sin a \log \mid \sin (x-a) \mid+c
Illustration:8. \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}
Solution: \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} \\ I=\int \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{\log x} d x} \\ =\int \frac{e^{\log x^5}-e^{\log x^4}}{e^{\log x^3}-e^{\log x^2}} d x \\ =\int \frac{x^5-x^4}{x^3-x^2} d x \\ =\int \frac{x^4(x-1)}{x^2(x-1)} d x \\ \Rightarrow I=\int x^2 d x \\ \Rightarrow I=\frac{1}{3} x^3+C
Illustration:9. \frac{\cos x}{\sqrt{4-\sin ^2 x}}
Solution: \frac{\cos x}{\sqrt{4-\sin ^2 x}} \\ I=\int \frac{\cos x}{\sqrt{4-\sin ^2 x}} d x Put \sin x=t \Rightarrow \cos x d x=d t \\ I=\int \frac{1}{\sqrt{4-t^2}} d t \\ =\sin ^{-1}\left(\frac{t}{2}\right)+C \\ \Rightarrow I=\sin ^{-1}\left(\frac{\sin x}{2}\right)+C
Illustration:10. \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x}
Solution: \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} \\ I=\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} \\ =\int \frac{\left(\sin ^4 x-\cos ^4 x\right)\left(\sin ^4 x+\cos ^4 x\right) d x}{\left(\sin ^2 x+\cos ^2 x\right)^2-2 \sin ^2 x \cos ^2 x} \\ =\int \frac{\left(\sin ^4 x-\cos ^4 x\right)\left(\sin ^4 x+\cos ^4 x\right) d x}{\left(\sin ^4 x+\cos ^4 x\right)+2 \sin ^2 x \cos ^2 x-2 \sin ^2 x \cos ^2 x} \\ =\int \frac{\left(\sin ^2 x-\cos ^2 x\right)\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^4 x+\cos ^4 x\right) d x}{\left(\sin^4 x+\cos ^4 x\right)} \\ =\int (1)\left(\sin ^2 x-\cos ^2 x\right) d x \\ =-\int \left(\cos ^2 x-\sin ^2 x\right) d x \\ =-\int \cos 2x dx \\ =-\frac{1}{2} \sin 2 x+C
Illustration:11. \frac{1}{\cos (x+a) \cos (x+b)}
Solution: \frac{1}{\cos (x+a) \cos (x+b)} \\ I=\int \frac{1}{\cos (x+a) \cos (x+b)} d x \\ =\frac{1}{\sin (a-b)} \int \frac{\sin (a-b)}{\cos (x+a) \cos (x+b)} d x \\ =\frac{1}{\sin (a-b)} \int \frac{\sin [(x+b)-(x+b)] d x}{\cos (x+a) \cos (x+b)} \\ =\frac{1}{\sin (a-b)} \int \frac{\sin (x+a) \cos (x+b)-\cos (x+a) \sin (x+b)}{\cos (x+a) \cos (x+b)} d x \\ =\frac{1}{\sin (a-b)} \int \frac{\sin (x+a) \cos (x+b)}{\cos (x+a) \cos (x+b)} d x -\frac{1}{\sin (a-b)} \int \frac{\cos (x+a) \sin (x+b)}{\cos (x+a) \cos (x+b)} d x \\ =\frac{1}{\sin (a-b)} \int \tan (x+a) d x-\frac{1}{\sin (a-b)} \int \tan (x+b) d x \\ =\frac{-1}{\sin (a-b)} \log |\cos (x+a)|+\frac{1}{\sin (a-b)} \log |\cos (x+b)|+c \\ =\frac{-1}{\sin (a-b)} \log |\cos (x+a)|+\frac{1}{\sin (a-b)} \log |\cos (x+b)|+c \\ \Rightarrow I=\frac{1}{\sin (a-b)} \log \left|\frac{\cos (x+b)}{\cos (x+a)}\right|+c
Illustration:12. \frac{x^3}{\sqrt{1-x^8}}
Solution: \frac{x^3}{\sqrt{1-x^8}} \\ I=\int \frac{x^3}{\sqrt{1-x^8}} d x \\ =\int \frac{x^3}{\sqrt{1-\left(x^4\right)^2}} dx
Put x^4=t \Rightarrow 4 x^3 d x=d t \\ \Rightarrow I=\frac{1}{4} \int \frac{1}{\sqrt{1-t^2}} d t \\ =\frac{1}{4} \sin ^{-1} t+C \\ \Rightarrow I=\frac{1}{4} \sin ^{-1}\left(x^4\right)+C

Indefinite Integrals in Class 12th

Illustration:13. \frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)}
Solution: \frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)} \\ I=\int \frac{e^x d x}{\left(1+ e^x \right) \left(2+e^x\right)}
Put e^x=t \Rightarrow e^x d x=d t \\ I=\int \frac{1}{(1+t)(t+2)} d t \\ =\int\left(\frac{1}{1+t}-\frac{1}{t+2}\right) d t \\ =\int \frac{1}{1+t} d t-\int \frac{1}{t+2} d t \\ =\log |1+t|-\log |t+2|+c \\ I=\log \left|\frac{1+t}{t+2}\right|+c \\ \Rightarrow I=\log \left|\frac{1+e^x}{2+e^x}\right|+C
Illustration:14. \frac{1}{\left(x^2+1\right)\left(x^2+4\right)}
Solution: \frac{1}{\left(x^2+1\right)\left(x^2+4\right)} \\ I=\int \frac{1}{\left(x^2+1\right) \left(x^2+4\right)} d x \\ =\frac{1}{3} \int\left(\frac{1}{x^2+1}-\frac{1}{x^2+4}\right) d x \\ =\frac{1}{3} \int \frac{1}{x^2+1} d x-\frac{1}{3} \int \frac{1}{x^2+4} d x \\ \Rightarrow I=\frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1}\left(\frac{x}{2}\right)+c
Illustration:15. \cos ^3 x e^{\log \sin x}
Solution: \cos ^3 x e^{\log \sin x} dx \\ I=\int \cos ^3 x e^{\log \sin x} d x \\ \Rightarrow I=\int \cos ^3 x \sin x dx
Put \cos x=t \Rightarrow-\sin x d x=d t \\ \Rightarrow I=-\int t^3 d t \\ =-\frac{1}{4} t^4+c \\ \Rightarrow I=-\frac{1}{4} \cos ^4 x+C
Illustration:16. e^{3 \log x} \left(x^4+1\right)^{-1}
Solution: e^{3 \log x} \left(x^4+1\right)^{-1} \\ I=\int e^{3 \log x}\left(x^4+1\right)^{-1} dx \\ =\int e^{\log x^3}\left(x^4+1\right)^{-1} dx \\=\int x^3\left(x^4+1\right)^{-1} dx \\ =\int \frac{x^3}{x^4+1} dx
Put x^4+1=t \\ 4 x^3 d x=d t \\ I=\frac{1}{4} \int \frac{1}{t} dt \\ =\frac{1}{4} \log t+C \\ I=\frac{1}{4} \log \left(x^4+1\right)+C
Illustration:17. f^{\prime}(a x+b)[f(a x+b)]^n
Solution: f^{\prime}(a x+b)[f(a x+b)]^n \\ I=\int f^{\prime}(a x+b)[f(a x+b)]^n dx
Put f(a x+b)=t \Rightarrow f^{\prime}(a x+b) \cdot a d x=d t \\ \Rightarrow I=\frac{1}{a} \int t^n d t \\ =\frac{1}{a} \cdot \frac{t^{n+1}}{n+1}+c \\ \Rightarrow I=\frac{1}{a(n+1)}[f(a x+b)]^{n+1}+c
Illustration:18. \frac{1}{\sqrt{\sin ^3 x \sin (x+\alpha)}}
Solution: \frac{1}{\sqrt{\sin ^3 x \sin (x+\alpha)}} \\ I=\int \frac{1}{\sqrt{\sin ^3 x \sin (x+\alpha)}} d x \\ =\int \frac{1}{\sqrt{\sin ^3 x(\sin x \cos \alpha+\cos x \sin \alpha)}} dx \\ =\int \frac{d x}{\sqrt{\sin ^4 x\left(\frac{\sin x \cos \alpha}{\sin x}+\frac{\cos x \sin \alpha}{\sin x}\right)}} \\ =\int \frac{d x }{\sin ^2 x \sqrt{\cos \alpha+\cot x \sin \alpha)}} \\ \Rightarrow I=\int \frac{\operatorname{cosec}^2 x d x}{\sqrt{\cos \alpha+\cot x \sin \alpha}}
Put \cos \alpha+\cot x \sin \alpha=t \\ \Rightarrow-\operatorname{cosec}^2 x \sin \alpha dx=dt \\ \Rightarrow \operatorname{cosec}^2 x d x=-\frac{1}{\sin \alpha} d t \\ \Rightarrow I=\frac{-1}{\sin \alpha} \int \frac{1}{\sqrt{t}} d t \\ =-\frac{1}{\sin \alpha} \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c \\ \Rightarrow I=-\frac{2}{\sin \alpha} \sqrt{\cos \alpha+\cot x \sin \alpha}+c \\ =-\frac{2}{\sin \alpha} \sqrt{\frac{\sin \alpha \cos \alpha+\cos x \sin \alpha}{\sin x}}+c \\ \Rightarrow I=-\frac{2}{\sin \alpha} \sqrt{\frac{\sin (x+\alpha)}{\sin x}}+c
Illustration:19. \frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}}(x \in[0,1])
Solution: \frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}} \\ I=\int \frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}} d x \\ =\int \frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\frac{\pi}{2}} d x \\ =\frac{2}{\pi} \int\left[\sin ^{-1} \sqrt{x}-\left(\frac{\pi}{2}-\sin \sqrt{x}\right)\right] dx \\ =\frac{2}{\pi} \int\left(\sin ^{-1} \sqrt{x}-\frac{\pi}{2}+\sin ^{-1} \sqrt{x}\right) d x \\ =\frac{4}{\pi} \int \sin ^{-1} \sqrt{x} d x-\int 1 \cdot d x
Put प्रथम समाकल में x=\sin ^2 t \Rightarrow d x=2 \sin t \cos t dt \\ \sin t=\sqrt{x}, \cos t=\sqrt{1-x} \\ =\frac{4}{\pi} \int \sin ^{-1} \sqrt{\sin ^2 t} \cdot 2 \sin t \cos t d t-x+C \\ =\frac{4}{\pi} \int t \sin 2 t-x+C \\ =\frac{4}{\pi} t \int \sin 2t d t-\frac{4}{\pi} \int\left[\frac{d}{d t}(t) \int \sin 2 t d t\right] dt-x+C \\ =\frac{4}{\pi} \quad t\left(-\frac{\cos 2 t}{2}\right)-\frac{4}{\pi} \int\left(-\frac{\cos 2 t}{2}\right) dt-x+C \\ =\frac{-2}{\pi} t \cos 2 t+\frac{1}{\pi} \sin 2 t-x+C \\ =-\frac{2}{\pi} \sin ^{-1} \sqrt{x}\left(1-2 \sin ^2 t\right)+\frac{2}{\pi} \sin t \cos t-x+C \\ =-\frac{2}{\pi} \sin^{-1} \sqrt{x}(1-2 x)+\frac{2}{\pi} \sqrt{x} \sqrt{1-x}-x+C \\ \Rightarrow I=\frac{2(2 x-1)}{\pi} \sin^{-1} \sqrt{x}+\frac{2}{\pi} \sqrt{x-x^2}-x+C
Illustration:20. \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}
Solution: \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} \\ I=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}} \times \frac{1-\sqrt{x}}{1-\sqrt{x}}} d x \\ =\int \sqrt{\frac{(1-\sqrt{x})^2}{1-x}} d x \\ =\int \frac{1-\sqrt{x}}{1-x} d x
Put x=\cos ^2 \theta \Rightarrow d x=-2 \cos \theta \sin \theta d \theta \\ =\int \frac{1-\cos \theta}{1-\cos ^2 \theta}(-2 \cos \theta \sin \theta) d \theta \\ =-2 \int \frac{1-\cos \theta}{\sin \theta}(\cos \theta \sin \theta) d \theta \\ =-2 \int(1-\cos \theta) \cos \theta d \theta \\ = -2 \int\left(\cos \theta-\cos ^2 \theta\right) d \theta \\ =-2 \int \cos \theta d \theta+2 \int \frac{1+\cos 2 \theta}{2} d \theta \\ =-2 \sin \theta+\int-1 \cdot d \theta+\int \cos 2 \theta d \theta \\ =-2 \sin \theta+\theta+\frac{1}{2} \sin 2 \theta+C \\ \Rightarrow I=-2 \sqrt{1-x}+\cos ^{-1} \sqrt{x}+\frac{1}{2} \times 2 \sin \theta \cos \theta+C \\ \Rightarrow I=-2 \sqrt{1-x}+\cos^{-1} \sqrt{x}+\sqrt{x-x^2}+C
Illustration:21. \frac{2+\sin 2 x}{1+\cos 2 x} e^x
Solution: \frac{2+\sin 2 x}{1+\cos } \cdot e^x \\ I=\int \frac{2+\sin 2 x}{1+\cos 2 x} e^x d x \\ =\int \frac{(2+2 \sin x \cos x)}{1+2 \cos ^2 x-1} e^x d x \\ =2 \int \frac{1+\sin x \cos x}{2 \cos ^2 x} e^x d x \\ =\int \frac{e^x}{\cos ^2 x} d x+\int e^x \tan x d x+c \\ =\int e^x\left(\sec ^2 x+\tan x\right) dx+C \\ f(x)=\tan x \Rightarrow f^{\prime}(x)=\sec ^2 x
अतः सूत्र I=\int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)  से:

\Rightarrow I=e^x \tan x+C
Illustration:22. \frac{x^2+x+1}{(x+1)^2(x+2)}
Solution: \frac{x^2+x+1}{(x+1)^2(x+2)} \\ I=\int \frac{x^2+x+1}{(x+1)^2(x+2)} d x
आंशिक भिन्नों में वियोजित करने पर:

\frac{x^2+x+1}{(x+1)^2(x+2)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+2} \\ \Rightarrow x^2+x+1=A(x+1)(x+2)+B(x+2)+C(x+1)^2
Put x=-1 \\ (-1)^2-1+1=B(-1+2) \Rightarrow B=1
Put x=-2 \\ (-2)^2-2+1=C(-2+1)^2 \Rightarrow C=3
Put x=0,1=2 A+2 B+C \\ 1=2 A+2+3 \Rightarrow A=-2 \\ I=\int \frac{-2}{x+1} d x+\int \frac{1}{(x+1)^2} d x+3 \int \frac{1}{x+2} dx \\ \Rightarrow I=-2 \log |x+1|-\frac{1}{(x+1)}+3 \log |x+2|+C
Illustration:23. \tan^{-1} \sqrt{\frac{1-x}{1+x}} 
Solution: \tan ^{-1} \sqrt{\frac{1-x}{1+x}} \\ I=\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} dx
Put x=\cos \theta \Rightarrow d x=-\sin \theta d \theta \\=\int \tan ^{-1} \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}(-\sin \theta) d \theta \\ =\int \tan \sqrt{\frac{2 \sin ^2 \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}}}(-\sin \theta) d \theta \\ =-\int \tan ^{-1}\left(\tan \frac{\theta}{2}\right) \sin \theta d \theta \\ =-\frac{1}{2} \int \theta \sin \theta d \theta \\ =\frac{1}{2} \theta \cos \theta-\frac{1}{2} \sin \theta+C \\ \Rightarrow I=\frac{1}{2}\left[x \cos ^{-1} x-\frac{1}{2} \sqrt{1-x^2}\right]+C
Illustration:24. \frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4}
Solution: \frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4} \\ I=\int \frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4} d x \\ =\int \frac{x \sqrt{1+\frac{1}{x^2}}}{x^4} \log \left(\frac{1+x^2}{x^2}\right) d x \\ =\int \frac{\sqrt{1+\frac{1}{x^2}}}{x^3} \log \left(1+\frac{1}{x^2}\right) dx
Put \frac{1}{x^2}=t^2 \Rightarrow-\frac{2}{x^3} dx=-2t dt \\ =-\int t^2 \log t^2 d t \\ =-2 \log t \int t^2 d t+2 \int\left[ \frac{d}{d t} \log t \int t^2 dt \right] dt \\ =-\frac{2 t^3}{3} \log t+2 \int \frac{1}{t} \cdot \frac{t^3}{3} d t+C \\ =-\frac{1}{3} t^3 \log t^2+\frac{2}{9} t^3+c \\ \Rightarrow I =-\frac{1}{3} \left(1+\frac{1}{x^2}\right)^{\frac{3}{2}}\left[\log \left(1+\frac{1}{x^2}\right)-\frac{2}{3}\right]+C
उपर्युक्त उदाहरणों के द्वारा कक्षा 12वीं अनिश्चित समाकल (Indefinite Integrals in Class 12th),अनिश्चित समाकलन कक्षा 12 (Indefinite Integral Class 12) को समझ सकते हैं।

Indefinite Integrals in Class 12th

Also Read This Article:- Definite Integrals by Substitution

3.कक्षा 12वीं अनिश्चित समाकल (Frequently Asked Questions Related to Indefinite Integrals in Class 12th),अनिश्चित समाकलन कक्षा 12 (Indefinite Integral Class 12) से सम्बन्धित अक्सर पूछे जाने वाले प्रश्न:

कक्षा 12वीं अनिश्चित समाकल (Indefinite Integrals in Class 12th) के इस आर्टिकल में
अनिश्चित समाकलनों पर आधारित कुछ विशिष्ट सवालों को हल करके समझने का प्रयास करेंगे।

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