Integral of $\arcsin$ of a rational function, using integration by parts

Question

I'm a class 12 student and this a question from my textbook:

$$I=\int{\arcsin{2x\over 1+x^2}}\mathrm{d}x$$

I did it using integration by parts like this:

$$I=\arcsin{\left(2x\over 1+x^2\right)}\cdot\int1\cdot\mathrm{d}x-\int(\frac{\mathrm{d}}{\mathrm{d}x}\arcsin{\left(2x\over1+x^2\right)}\int1\cdot\mathrm{d}x)\mathrm{d}x+c$$

Now,

$$\frac{\mathrm{d}}{\mathrm{d}x}\arcsin{\left(2x\over1+x^2\right)}\ = {1\over\sqrt{1-\left(2x\over1+x^2\right)^2}} { 2(1+x^2) - (2x)(2x)\over(1+x^2)^2}\ ={1+x^2(2(1-x^2)\over(1-x^2)^2(1+x^2)^2}\ ={2\over1-x^4}$$

So

$$I=\arcsin{\left(2x\over 1+x^2\right)}(x)-\int{2\over1-x^4}{x}\mathrm{d}x + c$$

let $I_1=\int{(2x)\over(1-x^4)}\mathrm{d}x$

$$I_1=\int{(2x)\over(1-x^2)(1+x^2)}\mathrm{d}x$$

Let $x^2=t$

So $2x\mathrm{d}x = \mathrm{d}t$ and

$$I_1=\int{\mathrm{d}t\over(1-t)(1+t)}\ =$\int{\mathrm{d}t\over1-t^2}\ ={1\over2}\log{|1+t|\over|1-t|}+c_2\ ={1\over2}\log{1+x^2\over1-x^2}+c_2$$

From all of this, we conclude

$$I=x \arcsin{\left(2x\over 1+x^2\right)}-{1\over2}\log{|1+x^2|\over|1-x^2|}+c$$

But the answer given in the book is :

$$(2x)\arctan x-{\log(1+x^2)} + c$$

I know they have done this using

$$\arcsin{\left(2x\over 1+x^2\right)}=2\arctan x$$

And then applied integration by parts, but I'd very much like to know where I went wrong.

Any help appreciated.

$\arcsin$ is another notation for the inverse sine function. Could you check that I did not introduce something wrong? Also, for your future LaTeX usage, you notice \frac{A}{B} is much easier than {A \over B}. — Jean-Claude Arbaut, Aug 24 '14 at 12:35
i asked because i was confused... thankyou@Jean-ClaudeArbaut — , Aug 24 '14 at 12:37
@kakashi, The last identity holds only if $x\le1$ See http://math.stackexchange.com/questions/523625/showing-arctan-frac23-frac12-arctan-frac125/523626#523626 — lab bhattacharjee, Aug 24 '14 at 14:34

score 0 · Accepted Answer · answered Aug 24 '14 at 12:44

HINT

Your calculation of $ \frac{\mathrm{d}}{\mathrm{d}x} \sin^{-1}\left (\frac{2x}{1+x^2} \right) $ is wrong. It will be:

$$\begin{align} \dfrac{\mathrm{d}}{\mathrm{d}x} \sin^{-1}\left (\dfrac{2x}{1+x^2} \right) &= {\dfrac{1}{\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}}} \cdot { 2(1+x^2) - (2x)(2x)\over(1+x^2)^2} \\ &= \color{red}{ \dfrac{2(1+x^2)(1-x^2)}{(1-x^2)(1+x^2)^2}} \\ &= \dfrac{2}{1+x^2} \end{align}$$

score 0 · Answer 2 · answered Aug 24 '14 at 13:24

$$\dfrac{\mathrm{d}}{\mathrm{d}x} \arcsin{\frac{2x}{1+x^2}}= {\dfrac{1}{\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}}} \cdot { 2(1+x^2) - (2x)(2x)\over(1+x^2)^2}=$$ $$=\color{red}{ \dfrac{2(1+x^2)(1-x^2)}{|1-x^2|(1+x^2)^2}} =$$ 1)$\color{red}{\dfrac{2}{1+x^2}}, x\in (-1,1)$

2)$\color{red}{-\dfrac{2}{1+x^2}}, x\in (-\infty,-1)\cup(1, \infty).$

Then integrating by parts, as you started, expressing full that the three intervals...

Integral of $\arcsin$ of a rational function, using integration by parts

2 Answers2

HINT