∫ {x x}{({x²-1})³} dx =

Correct Answer is : sec⁡−1x−log⁡xx2−1^{-1}x - { x}{{x²-1}}sec−1x−x2−1logx + c

Correct Answer is : sec⁡−1x−log⁡xx2−1^{-1}x - { x}{{x²-1}}sec−1x−x2−1logx + c

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AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 73 of 160

Marks: +1, -0

∫

\frac{x \log x}{(\sqrt{x^{2}-1})^{3}}

dx =

Solution:

Consider the integral,

I=\int \frac{x \log x}{(\sqrt{x^{2}-1})^{3}} dx

Let,

\sqrt{x^{2}-1}=t

\frac{2x dx}{2\sqrt{x^{2}-1}}=dt

\frac{x dx}{\sqrt{x^{2}-1}}=dt

So,

I=\int \frac{\log x dt}{t^{2}}

=\int \frac{\log (\sqrt{1+t^{2}})}{t^{2}} dt

=-\frac{1}{t} \log \sqrt{1+t^{2}}-\int \frac{2t}{2\sqrt{1+t^{2}}}

\left(-\frac{1}{t}\right)\frac{1}{\sqrt{1+t^{2}}} dt

Further simplify the above,

I=-\frac{1}{t} \log \sqrt{1+t^{2}}+\int \frac{dt}{1+t^{2}}

=-\frac{1}{t} \log \sqrt{1+t^{2}}+\tan^{-1}(t)+c

=\sec^{-1}(x)-\frac{\log x}{\sqrt{x^{2}-1}}+c

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