Solve Indefinite Integral of lnx/(1+x^2)^(3/2): Vuk's Q&A on Yahoo Answers

[MarkFL]

Here is the question:

How do you solve the integral of (lnx)dx/(1+x^2)^(3/2)?

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Vuk,

We are given to evaluate:

\(\displaystyle I=\int\frac{\ln(x)}{\left(x^2+1 \right)^{\frac{3}{2}}}\,dx\)

If we use integration by parts, we could let:

\(\displaystyle u=\ln(x)\,\therefore\,du=\frac{1}{x}\,dx\)

\(\displaystyle dv=\frac{1}{\left(x^2+1 \right)^{\frac{3}{2}}}\,dx\)

To find $v$, we may use a trigonometric substitution:

\(\displaystyle x=\tan(\theta)\,\therefore\,dx= \sec^2(\theta)\,d\theta\)

and we find:

\(\displaystyle v=\frac{\sec^2(\theta)}{\left(\tan^2(\theta)+1 \right)^{\frac{3}{2}}}\,d\theta\)

Using the Pythagorean identity \(\displaystyle \tan^2(\theta)+1=\sec^2(\theta)\) we get:

\(\displaystyle v=\frac{\sec^2(\theta)}{\sec^3(\theta)}\,d\theta= \int\cos(\theta)\,d\theta=\sin(\theta)\)

Back-substituting for $\theta$, we obtain:

\(\displaystyle v=\sin\left(\tan^{-1}(x) \right)=\frac{x}{\sqrt{x^2+1}}\)

And so we have:

\(\displaystyle I=\frac{x\ln(x)}{\sqrt{x^2+1}}-\int\frac{1}{\sqrt{x^2+1}}\,dx\)

Now, using the same trigonometric substitution we used before, we have:

\(\displaystyle I=\frac{x\ln(x)}{\sqrt{x^2+1}}- \int\sec(\theta)\,d\theta\)

Let:

\(\displaystyle u=\sec(\theta)+\tan(\theta)\, \therefore\,du= \left(\sec(\theta) \tan(\theta)+\sec^2( \theta) \right)\,d\theta=\)

\(\displaystyle \sec(\theta)\left(\tan(\theta)+\sec(\theta) \right)\,d\theta=u \sec(\theta)\,d\theta\,\therefore\,\sec(\theta)\,d\theta=\frac{1}{u}\,du\)

Hence, we now have:

\(\displaystyle I=\frac{x\ln(x)}{\sqrt{x^2+1}}-\int\frac{1}{u}\,du\)

\(\displaystyle I=\frac{x\ln(x)}{\sqrt{x^2+1}}-\ln|u|+C\)

Back-substitute for $u$:

\(\displaystyle I=\frac{x\ln(x)}{\sqrt{x^2+1}}-\ln|\sec(\theta)+\tan(\theta)|+C\)

Back-substitute for $\theta$:

\(\displaystyle I=\frac{x\ln(x)}{\sqrt{x^2+1}}-\ln\left|\sqrt{x^2+1}+x \right|+C\)

And we are done. (Sun)