1.कुछ विशिष्ट फलनों के समाकलन (Integrals of Some Particular Functions),कुछ विशिष्ट फलनों के समाकलन कक्षा 12 (Integrals of Some Particular Functions Class 12):

Integrals of Some Particular Functions

कुछ विशिष्ट फलनों के समाकलन (Integrals of Some Particular Functions) सूत्रों की सहायता से कुछ विशिष्ट प्रकार के 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:- Integration by Substitution in 12th

2.कुछ विशिष्ट फलनों के समाकलन के साधित उदाहरण (Integrals of Some Particular Functions Solved Examples):

प्रश्न 1 से 23 तक के फलनों का समाकलन कीजिए।
Example:1. \frac{3 x^2}{x^6+1}
Solution: \frac{3 x^2}{x^3+1} \\ I=\int \frac{3 x^2}{\left(x^3\right)^2+1} d x \\ \text{put } x^3=t \Rightarrow 3 x^2 d x=d t \\ =\int \frac{1}{t^2+1} d t \\ =\tan ^{-1} t+c \\ \Rightarrow I =\tan ^{-1}\left(x^3\right)+C
Example:2. \frac{1}{\sqrt{\left(1+4 x^2\right)}}
Solution: \frac{1}{\sqrt{\left(1+4 x^2\right)}} \\ I=\int \frac{1}{\sqrt{1+(2 x)^2}} d x \\ \text{put} 2 x=t \Rightarrow 2 d x=d t \\=\frac{1}{2} \int \frac{1}{\sqrt{1+t^2}} d t \\ =\frac{1}{2} \log \left|t+ \sqrt{1+t^2} \right|+C \\ \left[\because \int \frac{1}{\sqrt{x^2+a^2}} d x=\log \left|x+ \sqrt{x^2+a^2} \right| +c\right] \\ =\frac{1}{2} \log \left|2 x+\sqrt{1+4 x^2}\right|+C
Example:3. \frac{1}{\sqrt{(2-x)^2+1}}
Solution: \frac{1}{\sqrt{(2-x)^2+1}} \\ I =\int \frac{1}{\sqrt{(2-x)^2+1}} d x \\ =-\int \frac{1}{\sqrt{t^2+1}} d t \\ =-\log \left|t+\sqrt{t^2+1}\right|+C \\ =-\log \left|2-x+\sqrt{(2-x)^2+1}\right|+c \\ \Rightarrow I =\frac{1}{\log \left|2-x+\sqrt{(2-x)^2+1}\right|}+C
Example:4. \frac{1}{\sqrt{9-25 x^2}}
Solution: \frac{1}{\sqrt{9-25 x^2}} \\ I=\int \frac{1}{\sqrt{9-25 x^2}} d x \\ \text{put } 5 x=t \Rightarrow 5 d x=d t \\ =\frac{1}{5} \int \frac{1}{\sqrt{3^2-t^2}} d t \\ =\frac{1}{5} \sin^{-1} \left(\frac{t}{3}\right)+c \\ \left[\because \frac{1}{\sqrt{a^2-x^2}} d x=\sin ^{-1} \frac{x}{a}\right] \\ \Rightarrow I=\frac{1}{5} \sin^{-1} \left(\frac{5 x}{3}\right)+c
Example:5. \frac{3 x}{1+2 x^4}
Solution: \frac{3 x}{1+2 x^4} \\ I=\int \frac{3 x}{1+2 x^4} d x \\ I=\int \frac{3 x}{1+\left(\sqrt{2} x^2\right)^2} dx \\ \text { Put } \sqrt{2} x^2=t \Rightarrow 2 \sqrt{2 x}d x=d t \\ I=\frac{1}{2 \sqrt{2}} \int \frac{3 d t}{\sqrt{1+t^2}} \\ =\frac{3}{2 \sqrt{2}} \tan^{-1} t+C \\ =\frac{3}{2 \sqrt{2}} \tan^{-1} \left(\sqrt{2} x^2\right)
Example:6. \frac{x^2}{1-x^6}
Solution: \frac{x^2}{1-x^6} \\ I=\int \frac{x^2}{1-x^6} d x \\ =\int \frac{x^2}{1-\left(x^3\right)^2} d x \\ \text { put } x^3=t \Rightarrow 3 x^2 d x=d t \\ =\frac{1}{3} \int \frac{1}{1-t^2} d t \\ =\frac{1}{3} \cdot \frac{1}{2} \log \left|\frac{1+t}{1-t}\right|+C \\ \left[\because \int \frac{d x}{a^2-x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+c\right] \\ \Rightarrow I=\frac{1}{6} \log \left|\frac{1+x^3}{1-x^3}\right| +C
Example:7. \frac{x-1}{\sqrt{x^2-1}}
Solution: \frac{x-1}{\sqrt{x^2-1}} \\ I=\int \frac{x-1}{\sqrt{x^2-1}} d x \\ =\int \frac{x}{\sqrt{x^2-1}} d x-\int \frac{1}{\sqrt{x^2-1}} d x
Put प्रथम समाकल में:

x^2-1=t \Rightarrow 2 x d x=d t \\ =\frac{1}{2} \int \frac{1}{\sqrt{t}} d t-\log \left|x+\sqrt{x^2-1}\right|+c \\ =\frac{1}{2} \times 2 \sqrt{t}-\log \left|x+\sqrt{x^2-1}\right|+c \\ =\sqrt{x^2-1}-\log |x+\sqrt{x^2-1}|+c
Example:8. \frac{x^2}{\sqrt{x^6+a^6}}
Solution: \frac{x^2}{\sqrt{x^6+a^6}} \\ I =\int \frac{x^2}{\sqrt{x^6+a^6}} d x \\ =\int \frac{x^2}{\sqrt{\left(x^3\right)^2+a^6}} d x \\ =\text{put } x^3=t \Rightarrow 3 x^2 d x=d t \\ =\frac{1}{3}\int \frac{1}{\sqrt{t^2+\left(a^3\right)^2}} d t \\ =\frac{1}{3} \log \left|t+\sqrt{t^2+a^6}\right|+C \\ \Rightarrow I=\frac{1}{3} \log \left|x^3+\sqrt{x^6+a^6}\right|+C
Example:9. \frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}
Solution: \frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}} \\ I=\int \frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}} d x \\ \text{Put } \tan x=t \Rightarrow \sec ^2 x d x=d t \\ =\int \frac{1}{\sqrt{t^2+4}} d t \\ =\log \left|t+\sqrt{t^2+4} \right|+c \\ \Rightarrow I =\log \left|\tan x+\sqrt{\tan ^2 x+4}\right|+c
Example:10. \frac{1}{\sqrt{x^2+2 x+2}}
Solution: \frac{1}{\sqrt{x^2+2 x+2}} \\ I=\int \frac{1}{\sqrt{x^2+2 x+2}} d x \\ =\int \frac{d x}{\sqrt{\left(x+1\right)^2+1^2}} \\ \text { Put } x+1=t \Rightarrow d x=d t \\ I=\int \frac{d t}{\sqrt{t^2+1}} \\ =\log \left|t+\sqrt{t^2+1}\right|+C \\ \Rightarrow I=\log \left|x+1+\sqrt{x^2+2 x+1}\right|+C
Example:11. \frac{1}{9 x^2+6 x+5}
Solution: \frac{1}{9 x^2+6 x+5} \\ I=\int \frac{1}{9 x^2+6 x+5} d x \\ =\int \frac{d x}{(3 x)^2+6 x+1+4} \\ =\int \frac{d x}{(3 x+1)^2+2^2} \\ \text { Put } 3 x+1=t \Rightarrow 3 d x=d t \\ I=\frac{1}{3} \int \frac{d t}{t^2+2^2} \\ =\frac{1}{6} \tan^{-1} \left(\frac{t}{2}\right)+c\\ =\frac{1}{6} \tan ^{-1}\left(\frac{3 x +1}{2}\right)+C
Example:12. \frac{1}{\sqrt{7-6 x-x^2}}
Solution: \frac{1}{\sqrt{7-6 x-x^2}} \\ I=\int \frac{d x}{\sqrt{7-6 x-x^2}} \\ =\int \frac{d x}{\sqrt{7-\left(6 x+x^2\right)}} \\ =\int \frac{d x}{\sqrt{9+7-\left(6 x+x^2+9\right)}} \\ =\int \frac{d x}{\sqrt{16-(x+3)^2}} \\ \text{put } x+3=t \Rightarrow d x=d t \\ I=\int \frac{d t}{\sqrt{4^2-t^2}} \\ =\sin ^{-1}\left(\frac{t}{4}\right)+ \\ \Rightarrow I=\sin^{-1} \left(\frac{x+3}{4}\right)+C
Example:13. \frac{1}{\sqrt{(x-1)(x-2)}}
Solution: \frac{1}{\sqrt{(x-1)(x-2)}} \\ I=\int \frac{d x}{\sqrt{(x-1)(x-2)}} \\ =\int \frac{d x}{\sqrt{x^2-3 x+2}} \\ I=\int \frac{d x}{\sqrt{(x-1)(x-2)}} \\ =\int \frac{d x}{\sqrt{x^2-3 x+2}} \\ =\int \frac{d x}{\sqrt{x^2-\frac{3 x}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+2}} \\ =\int \frac{d x}{\sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}} \\ \text { put } x-\frac{3}{2}=t \Rightarrow d x=d t \\ I=\int \frac{1}{\sqrt{t^2-\left(\frac{1}{2}\right)^2}} \\ =\log \left|t+\sqrt{t^2-\left(\frac{1}{2}\right)^2}\right| t c \\ \Rightarrow I=\log \left|x-\frac{3}{2}+\sqrt{x^2-3 x+2}\right|+c

Integrals of Some Particular Functions

Example:14. \frac{1}{\sqrt{8+3 x-x^2}}
Solution: \frac{1}{\sqrt{8+3 x-x^2}} \\ I=\int \frac{d x}{\sqrt{8+3 x-x^2}} \\ =\int \frac{d x}{\sqrt{8-\left(x^2-3 x\right)}} \\ =\int \frac{d x}{\sqrt{8+\left(\frac{3}{2}\right)^2-\left(x^2-3 x+\left(\frac{3}{2}\right)^2 \right)}} \\ =\int \frac{d x}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2-\left(x-\frac{3}{2}\right)^2}} \\ \text{put } x-\frac{3}{2}=t \Rightarrow d x=d t \\ I=\int \frac{d t}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2-t^2}} \\ =\sin ^{-1}\left(\frac{t}{\frac{\sqrt{41}}{2}}\right)+c \\ =\sin ^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\right)+c \\ \Rightarrow I =\sin ^{-1}\left(\frac{2 x-3}{\sqrt{41}}\right)+c
Example:15. \frac{1}{\sqrt{(x-a)(x-b)}}
Solution: \frac{1}{\sqrt{(x-a)(x-b)}} \\ I=\int \frac{d x}{\sqrt{(x-a)(x-b)}} \\ =\int \frac{d x}{\sqrt{x^2-(a+b) x+a b}} \\ =\int \frac{d x}{\sqrt{x^2-(a+b) x+\left(\frac{a+b}{2}\right)^2+a b-\left(\frac{a+b}{2}\right)^2}} \\ =\int \frac{d x}{\sqrt{\left(x-\frac{a+b}{2}\right)^2-\left(\frac{\sqrt{a^2+b^2}}{2}\right)^2}} \\ \text{Put } x-\frac{a+b}{2}=t \Rightarrow d x=d t \\ I=\int \frac{d t}{\sqrt{t^2-\left(\frac{\sqrt{a^2+b^2}}{2}\right)^2}} \\ =\log \left\lvert\, t+\sqrt{t^2-\left(\frac{\sqrt{a^2+b^2}}{2}\right)^2 } \right\rvert\,+c \\ =\log \left|x-\frac{a+b}{2}+\sqrt{x^2-(a+b) x+a b}\right|+c \\ \Rightarrow I=\log \left|x-\frac{a+b}{2} +\sqrt{(x-a)(x-b)} \right\rvert\,+c
Example:16. \frac{4 x+1}{\sqrt{2 x^2+x-3}}
Solution: \frac{4 x+1}{\sqrt{2 x^2+x-3}} \\ I=\int \frac{4 x+1}{\sqrt{\left(2 x^2+x-3\right)}} d x \\ \text{Put } 2 x^2+x-3=t \Rightarrow(4 x+1) d x=dt \\ =\int \frac{d t}{\sqrt{t}} \\ =2 \sqrt{t}+C \\ \Rightarrow I=2 \sqrt{\left(2 x^2+x-3\right)}+C
Example:17. \frac{x+2}{\sqrt{x^2-1}}
Solution: \frac{x+2}{\sqrt{x^2-1}} \\ I=\int \frac{x+2}{\sqrt{x^2-1}} d x \\ =2 \int \frac{2 x}{\sqrt{x^2-1}} d x+\int \frac{2}{\sqrt{x^2-1}} d x
Put प्रथम समाकल में: x^2-1=t \Rightarrow 2 x=d t \\ =2 \int \frac{d t}{\sqrt{t}}+2 \log \left|x+ \sqrt{x^2-1}\right|+c \\ =2 \times 2 \sqrt{t}+2 \log \left|x+\sqrt{x^2-1}\right|+c \\ \Rightarrow I=4 \sqrt{x^2-1}+2 \log \left|x+\sqrt{x^2-1}\right|+C
Example:18. \frac{5 x-2}{1+2 x+3 x^2}
Solution: \frac{5 x-2}{1+2 x+3 x^2} \\ I=\int \frac{(5 x-2) d x}{1+2 x+3 x^2} \\ 5 x-2=A \frac{d}{d x}\left(1+2 x+3 x^2\right)+B \\ =A(6 x+2)+B \\ \Rightarrow 5 x-2=6 A x+2 A+B
तुलना करने पर:

6A =5 \Rightarrow A=\frac{5}{6} \\ 2 A+B =-2 \Rightarrow 2 \times \frac{5}{6}+B=-2 \\ \Rightarrow B =-2-\frac{5}{3} \\ \Rightarrow B =-\frac{11}{3} \\ I =\int \frac{\frac{5}{6}(6 x+2)-\frac{11}{3}}{1+2 x+3 x^2} d x \\ =\frac{5}{6} \int \frac{6 x+2}{1+2 x+3 x^2} d x-\frac{11}{3} \int \frac{d x}{1+2 x+3 x^2}
Put प्रथम समाकल में:

3 x^2+2 x+1=t \Rightarrow(6 x+2) d x=d t \\ =\frac{5}{6} \int \frac{1}{t} d t-\frac{11}{3} \int \frac{d x}{(\sqrt{3} x)^2+2 x+(\frac{1}{\sqrt{3}})^2-\frac{1}{3}+1} \\=\frac{5}{6} \log |t|-\frac{11}{3} \int \frac{d x}{\left(\sqrt{3} x+\frac{1}{\sqrt{3}}\right)^2+\left(\sqrt{\frac{2}{3}}\right)^2}
Put द्वितीय समाकल में:

\sqrt{3} x+\frac{1}{\sqrt{3}}=u \Rightarrow \sqrt{3} d x=d u \\ =\frac{5}{6} \log \left|3 x^2+2 x+1\right|-\frac{11}{3 \sqrt{3}} \int \frac{d t}{t^2+\left(\frac{\sqrt{2}}{3}\right)^2} \\ =\frac{5}{6} \log |3 x^2+2 x+1|-\frac{11}{3 \sqrt{3}} \cdot \frac{1}{\left(\sqrt{\frac{2}{3}}\right)} \tan^{-1} \left(\frac{t}{\left(\sqrt{\frac{2}{3}}\right)}\right)+c \\ =\frac{5}{6} \log \left|3 x^2+2 x+1\right|-\frac{11}{3 \sqrt{2}} \tan ^{-1} \left(\frac{\sqrt{3} x+\frac{1}{\sqrt{3}}}{\sqrt{\frac{2}{3}}}\right)+c \\ \Rightarrow I=\frac{5}{6} \log \left|3 x^2+2 x+1\right|-\frac{11}{3 \sqrt{2}} \tan^{-1} \left(\frac{3 x+1}{\sqrt{2}}\right)+c
Example:19. \frac{6 x+7}{\sqrt{(x-5)(x-4)}}
Solution: I=\int \frac{6 x+7}{\sqrt{(x-5)(x-4)}} d x \\ =\int \frac{6 x+7}{\sqrt{x^2-9 x+20}} d x \\ 6 x+7 =A \frac{d}{d x}\left(x^2-9 x+20\right)+B \\ =A(2 x-9)+B \\ \Rightarrow 6 x+7=2 A x-9 A+B
तुलना करने पर:

2 A=6 \Rightarrow A=3 \\ -9 A+B=7 \Rightarrow-9 \times 3+B=7 \\ \Rightarrow B=34 \\ I=\int \frac{A \frac{d}{d x}\left(x^2-9 x+20\right)+B}{\sqrt{x^2-9 x+20}} d x \\ =\int \frac{3(2 x-9)}{\sqrt{x^2-9 x+20}} d x+34 \int \frac{1}{\sqrt{x^2-9 x+20}} d x
Put प्रथम समाकल में:

x^2-9 x+20=t \Rightarrow(2 x-9) d x=d t \\ =3 \int \frac{1}{\sqrt{t}} d t+34 \int \frac{d x}{\sqrt{\left(x-\frac{9}{2}\right)^2+20-\frac{81}{4}}} \\ =3 \times 2 \sqrt{t}+34 \int \frac{d x}{\sqrt{\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2}}
Put द्वितीय समाकल में:

x-\frac{9}{2}=u \Rightarrow d x=d u \\ = 6 \sqrt{x^2-9 x+20}+34 \int \frac{d u}{\sqrt{u^2-\left(\frac{1}{2}\right)^2}} \\ =6 \sqrt{x^2-9 x+20}+34 \log \left|u+\sqrt{u^2-\left(\frac{1}{2}\right)^2}\right|+c \\ \Rightarrow I=6 \sqrt{x^2-9 x+20}+34 \log \left\lvert\, x-\frac{9}{2}+ \sqrt{\left(x-\frac{9}{2} \right)^2-\left(\frac{1}{2}\right)^2}\right|+c \\ \Rightarrow I=6 \sqrt{x^2-9 x+20}+34 \log \left|x-\frac{9}{2}+\sqrt{x^2-9 x+20} \right|+C
Example:20. \frac{x+2}{\sqrt{4 x-x^2}}
Solution: \frac{x+2}{\sqrt{4 x-x^2}} \\ I=\int \frac{x+2}{\sqrt{4 x-x^2}} d x \\ x+2=A \frac{d}{d x}\left(4 x-x^2\right)+B \\ =A(4-2 x)+B \\ \Rightarrow x+2=-2 A x+4 A+B
तुलना करने पर:

-2 A=1 \Rightarrow A=-\frac{1}{2} \\ \\ 4 A+B=2 \Rightarrow 4 x-\frac{1}{2}+B=2 \\ \Rightarrow B=4 \\ I=\int \frac{A \frac{d}{d x}\left(4 x-x^2\right)+B}{\sqrt{4 x-x^2}} d x \\ =-\frac{1}{2} \int \frac{(4-2 x)}{\sqrt{4 x-x^2}} d x+\frac{1}{4} \int \frac{dx}{\sqrt{4x-x^2}}
Put प्रथम समाकल में:

4 x-x^2=t \Rightarrow(4-2 x) d x=d t \\ I=\frac{-1}{2} \int \frac{1}{\sqrt{t}}+4 \int \frac{d x}{\sqrt{4-\left(x^2-4 x+4\right)}} \\ =-\frac{1}{2} \times 2 \sqrt{t}+4 \int \frac{d x}{\sqrt{2^2-(x-2)^2}} \\ =-\sqrt{4 x-x^2}+4 \sin^{-1} \left(\frac{x-2}{2}\right)+c \\ \Rightarrow I=-\sqrt{4 x-x^2}+4 \sin ^{-1}\left(\frac{x-2}{2}\right)+c
Example:21. \frac{x+2}{\sqrt{x^2+2 x+3}}
Solution: \frac{x+2}{\sqrt{x^2+2 x+3}} \\ I=\int \frac{x+2}{\sqrt{x^2+2 x+3}} d x \\ x+2=A \frac{d}{d x}\left(x^2+2 x+3\right)+B \\=A(2 x+2)+B \\ \Rightarrow x+2=2 A x+2 A+B
तुलना करने पर:

2 A=1 \Rightarrow A=\frac{1}{2} \\ 2 A+B=2 \Rightarrow 2 \times \frac{1}{2}+B=2 \Rightarrow B=1 \\ I=\frac{1}{2} \int \frac{2 x+2}{\sqrt{x^2+2 x+3}} d x+\int \frac{d x}{\sqrt{x^2+2 x+3}} \\ =\frac{1}{2} \times 2 \sqrt{x^2+2 x+3}+\int \frac{d x}{\sqrt{\left(x+1\right)^2+(\sqrt{2})^2}} \\ \Rightarrow I=\sqrt{x^2+2 x+3}+\log \mid x+1+\sqrt{x^2+2 x+3}\mid+C
Example:22. \frac{x+3}{x^2-2 x-5}
Solution:\frac{x+3}{x^2-2 x-5} \\ I=\int \frac{x+3}{x^2-2 x-5} d x \\ x+3=A \frac{d}{d x}\left(x^2-2 x-5\right)+B \\ \Rightarrow x+3=A(2 x-2)+B \\ \Rightarrow x+3=2 A x-2 A+B
तुलना करने पर:

2 A=1 \Rightarrow A=\frac{1}{2},-2 A+B=3 \\ -2 \times \frac{1}{2}+B=3 \Rightarrow B=4 \\ I=\frac{1}{2} \int \frac{(2 x-2)}{x^2-2 x-5} d x+4 \int \frac{1}{x^2-2 x-5} d x \\ =\frac{1}{2} \log \left|x^2-2 x-5\right|+4 \int \frac{d x}{(x-1)^2-(\sqrt{6})^2} \\ \Rightarrow I=\frac{1}{2} \log \left|x^2-2 x-5\right|+\frac{4}{2\sqrt{6}} \log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|+C \\ \Rightarrow I=\frac{1}{2} \log \left|x^2-2 x-5\right|+ \frac{2}{\sqrt{6}} \log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|+C
Example:23. \frac{5 x+3}{\sqrt{x^2+4 x+10}}
Solution: \frac{5 x+3}{\sqrt{x^2+4 x+10}} \\ I=\int \frac{5 x+3}{\sqrt{x^2+4 x+10}} \\ 5 x+3=A \frac{d}{d x}\left(x^2+4 x+10\right)+B \\ =A(2 x+4)+B \\ \Rightarrow 5 x+3=2 A x+4 A+B
तुलना करने पर:

2 A=5 \Rightarrow A=\frac{5}{2} \\ 4 A+B=3 \Rightarrow 4 \times \frac{5}{2}+B=3 \Rightarrow B=-7 \\ I= \frac{-5}{2} \int \frac{(2 x+4)}{\sqrt{x^2+4 x+10}} d x-7 \int \frac{d x}{\sqrt{x^2+4x+10}} \\ =\frac{5}{2} \int \frac{1}{\sqrt{t}} d t-7 \int \frac{d x}{\sqrt{(x+2)^2+(\sqrt{6})^2}} \\ =\frac{5}{2} \times 2 \sqrt{t}-7 \log \left\lvert\, x+2+\sqrt{x^2+4 x+10} \right\rvert\,+C \\ \Rightarrow I =5 \sqrt{x^2+4 x+10} -7 \log \left\lvert\, x+2+\sqrt{x^2+4 x+10}\right\rvert\, +C
प्रश्न 24 एवं 25 में सही उत्तर का चयन कीजिए:
Example:24. \frac{d x}{x^2+2 x+2} बराबर है:

(A) x \tan ^{-1}(x+1)+c        (B) \tan^{-1} (x+1)+C
(C) (x+1) \tan ^{-1} x+C       (D) \tan ^{-1} x+C
Solution: \int \frac{d x}{x^2+2 x+2} d x \\ =\int \frac{d x}{(x+1)^2+1^2} \\ =\tan ^{-1}(x+1)+C
अतः विकल्प (B) सही है।
Example:25. \frac{d x}{\sqrt{9 x-4 x^2}} बराबर है:

(A) \frac{1}{9} \sin^{-1} \left(\frac{9 x-8}{8}\right)+C
(B) \frac{1}{2} \sin ^{-1}\left(\frac{8 x-9}{9}\right)+C
(C) \frac{1}{3} \sin^{-1} \left(\frac{9 x-8}{8}\right)+C
(D) \frac{1}{2} \sin ^{-1}\left(\frac{9 x-8}{9}\right)+C
Solution: \int \frac{d x}{\sqrt{9 x-4 x^2}} \\ =\int \frac{d x}{\sqrt{\left(\frac{9}{4}\right)^2-\left(4 x^2-9 x+\left(\frac{9}{4}\right)^2\right)}} \\ =\int \frac{d x}{\sqrt{\left(\frac{9}{4}\right)^2-\left(2 x-\frac{9}{4}\right)^2}} \\ \text{put } 2 x-\frac{9}{4}=t \Rightarrow 2 d x=d t \\ =\frac{1}{2} \int \frac{d t}{\sqrt{\left(\frac{9}{4}\right)^2-t^2}} \\ =\frac{1}{2} \int \frac{d t}{\sqrt{\left(\frac{9}{4}\right)^2-t^2}} \\ =\frac{1}{2} \sin ^{-1}\left(\frac{t}{\frac{9}{4}}\right)+c \\ =\frac{1}{2} \sin ^{-1}\left(\frac{2 x-\frac{9}{4}}{\frac{9}{4}}\right)+c \\ =\frac{1}{2} \sin ^{-1}\left(\frac{8 x-9}{9}\right)+c
विकल्प (B) सही है।
उपर्युक्त उदाहरणों के द्वारा कुछ विशिष्ट फलनों के समाकलन (Integrals of Some Particular Functions),कुछ विशिष्ट फलनों के समाकलन कक्षा 12 (Integrals of Some Particular Functions Class 12) को समझ सकते हैं।

3.कुछ विशिष्ट फलनों के समाकलन के सवाल (Integrals of Some Particular Functions Questions):

Integrals of Some Particular Functions

निम्न का x के सापेक्ष समाकलन कीजिए:

(1.) \frac{x^3}{\sqrt{1-x^8}} (2.) \frac{1}{x \sqrt{x^4-1}}
उत्तर (Answers):(1.) \frac{1}{4} \sin^{-1} x^4+c
(2.) \frac{1}{2} \sec^{-1} x^2+c
उपर्युक्त सवालों को हल करने पर कुछ विशिष्ट फलनों के समाकलन (Integrals of Some Particular Functions),कुछ विशिष्ट फलनों के समाकलन कक्षा 12 (Integrals of Some Particular Functions Class 12) को ठीक से समझ सकते हैं।

Also Read This Article:- Integration Class 12

4.कुछ विशिष्ट फलनों के समाकलन (Frequently Asked Questions Related to Integrals of Some Particular Functions),कुछ विशिष्ट फलनों के समाकलन कक्षा 12 (Integrals of Some Particular Functions Class 12) से सम्बन्धित अक्सर पूछे जाने वाले प्रश्न:

कुछ विशिष्ट फलनों के समाकलन (Integrals of Some Particular Functions) सूत्रों
की सहायता से कुछ विशिष्ट प्रकार के सवालों को हल करके समझने का प्रयास करेंगे।

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