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NCERT Class XII Mathematics Chapter - - Solutions

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..... Miscellaneous Exercise .....
Integrate the functions in Exercises 229 to 252.
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Question : 252 of 271
Marks: +1, -0
x2+1[log⁡(x2+1)−2log⁡x]x4\frac{\sqrt{x^2+1}[\log(x^2+1)-2\log x]}{x^4}
Solution:  
Let I=∫x2+1[log⁡(x2+1)−2log⁡x]x4dxI=\int\frac{\sqrt{x^2+1}[\log(x^2+1)-2\log x]}{x^4}dx
=∫x2+1x2log⁡(x2+1x2)1x3dx=\int\sqrt{\frac{x^2+1}{x^2}}\log\left(\frac{x^2+1}{x^2}\right)\frac{1}{x^3}dx
=∫1+1x2log⁡(1+1x2)1x3dx=\int\sqrt{1+\frac{1}{x^2}}\log\left(1+\frac{1}{x^2}\right)\frac{1}{x^3}dx
Put 1x2=t⇒x−2=t\frac{1}{x^2}=t\Rightarrow x^{-2}=t
⇒−2x−3dx=dt\Rightarrow -2x^{-3}dx=dt
⇒−2x3dx=dt\Rightarrow -\frac{2}{x^3}dx=dt
∴I=−12∫1+tlog⁡(1+t)dt\therefore I=-\frac{1}{2}\int\sqrt{1+t}\log(1+t)dt
∴I=−12[log⁡(1+t)⋅(1+t)3/232−∫11+t(1+t)3/232dt]\therefore I=-\frac{1}{2}\left[\log(1+t)\cdot\frac{(1+t)^{3/2}}{\frac{3}{2}}-\int\frac{1}{1+t}\frac{(1+t)^{3/2}}{\frac{3}{2}}dt\right]
=−12[23(1+t)3/2log⁡(1+t)−23∫(1+t)1/2dt]=-\frac{1}{2}\left[\frac{2}{3}(1+t)^{3/2}\log(1+t)-\frac{2}{3}\int(1+t)^{1/2}dt\right]
=−12[23(1+t)3/2log⁡(1+t)−23(1+t)3/232]+C=-\frac{1}{2}\left[\frac{2}{3}(1+t)^{3/2}\log(1+t)-\frac{2}{3}\frac{(1+t)^{3/2}}{\frac{3}{2}}\right]+C
=−12[23(1+t)3/2log⁡(1+t)−49(1+t)3/2]+C=-\frac{1}{2}\left[\frac{2}{3}(1+t)^{3/2}\log(1+t)-\frac{4}{9}(1+t)^{3/2}\right]+C
=−13(1+t)3/2log⁡(1+t)+29(1+t)3/2+C=-\frac{1}{3}(1+t)^{3/2}\log(1+t)+\frac{2}{9}(1+t)^{3/2}+C
=−13(1+1x2)3/2log⁡(1+1x2)+29(1+1x2)3/2+C=-\frac{1}{3}\left(1+\frac{1}{x^2}\right)^{3/2}\log\left(1+\frac{1}{x^2}\right)+\frac{2}{9}\left(1+\frac{1}{x^2}\right)^{3/2}+C
=−13[(1+1x2)3/2(log⁡(1+1x2)−23)]+C=-\frac{1}{3}\left[\left(1+\frac{1}{x^2}\right)^{3/2}\left(\log\left(1+\frac{1}{x^2}\right)-\frac{2}{3}\right)\right]+C
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