e^{-x}^{-1}(e^x)dx=f(x)-1/2(1+e²x})+c f(x)=

Correct Answer is : x−e−xtan⁡−1(ex)x-e^{-x}^{-1}(e^x)x−e−xtan−1(ex)

Correct Answer is : x−e−xtan⁡−1(ex)x-e^{-x}^{-1}(e^x)x−e−xtan−1(ex)

Test Index

TS EAMCET 2015 Solved Paper

Section: Mathematics

Question : 9 of 160

Marks: +1, -0

\int e^{-x}\tan^{-1}(e^x)dx=f(x)-\frac{1}{2}\log(1+e^{2x})+c\Rightarrow f(x)=

Solution:

Put

e^x=t, e^xdx=dt, dx=\frac{1}{t}dt

=\int \frac{\tan^{-1}}{t^2}dt

=\tan^{-1}t\left(-\frac{1}{t}\right)-\int \left(-\frac{1}{t}\right)\frac{1}{1+t^2}dt

=\frac{-\tan^{-1}t}{t}+\frac{1}{2}\int \frac{2t}{t^2(1+t^2)}dt

t^2=z, 2t dt=dz

=\frac{-\tan^{-1}t}{t}+\frac{1}{2}\int \frac{(z+1)-z}{z(1+z)}dz

=\frac{-\tan^{-1}t}{t}+\frac{1}{2}\left[\int \frac{1}{z}dz-\int \frac{1}{z+1}dz\right]

=\frac{-\tan^{-1}t}{t}+\frac{1}{2}\log\left|\frac{t^2}{1+t^2}\right|+c

=\frac{-\tan^{-1}t}{t}+\frac{1}{2}\log\left|\frac{e^{2x}}{1+e^{2x}}\right|+c

=-e^{-x}\tan^{-1}(e^x)-\frac{1}{2}\log\left|1+e^{2x}\right|+\frac{1}{2}\cdot 2x+c

=x-e^{-x}\tan^{-1}(e^{x})-\frac{1}{2}\log\left|1+e^{x}\right|+c

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