Integrate $\int \ln(x^2 +1)\ dx$

Question

$$\int \ln(x^2 +1)\ dx$$

I done it using integration by parts where

$\int u\ dv = uv - \int v\ du$

Let $u$ = $\ln(x^2 +1)$

$du = \frac{2x}{x^2+1} dx $

Let $dv = dx$ so $v=x$

$\int \ln(x^2 +1)\ dx = x \ln (x^2 +1) - \int \frac{2x^2}{x^2+1} $

I integrate $2\int \frac{x^2}{x^2+1} $ separately using substitution.

Let $u$ = $x^2 + 1$ -> $dx= \frac{1}{2x} du $

substitute $u$ back into it I get a simplified $\int \frac{x}{x^2 + 1} dx$

I used substitution again and got $ \frac{1}{2} \int \frac{1}{u} du$

When I sub it back into the original question, this is my final answer,

$x\ln(x^2 +1) + \frac{1}{2} \ln(x^2 +1) + C$

However, this answer is wrong,

The answer is, $x\ln(x^2 +1) - 2x + 2\tan^{-1} (x) + C $

I believe my integration of $2\int \frac{x^2}{x^2+1} $ is wrong. Where did I went wrong ?

Answer 1

Your integration by parts in the beginning is correct. Truly :

$${\displaystyle\int}\ln\left(x^2+1\right)\,\mathrm{d}x =x\ln\left(x^2+1\right)-{\displaystyle\int}\dfrac{2x^2}{x^2+1}\,\mathrm{d}x$$

Now, handling the second integral, first of all let's factor out the constant and write $x^2$ as $x^2 + 1 - 1$ to split it up :

$${\displaystyle\int}\dfrac{2x^2}{x^2+1}\,\mathrm{d}x =\class{steps-node}{\cssId{steps-node-1}{2}}{\displaystyle\int}\dfrac{x^2}{x^2+1}\,\mathrm{d}x ={\displaystyle\int}\left(\dfrac{\class{steps-node}{\cssId{steps-node-4}{x^2+1}}}{x^2+1}-\dfrac{\class{steps-node}{\cssId{steps-node-5}{1}}}{x^2+1}\right)\mathrm{d}x $$

$$=$$

$$={\displaystyle\int}\left(1-\dfrac{1}{x^2+1}\right)\mathrm{d}x ={\displaystyle\int}1\,\mathrm{d}x-{\displaystyle\int}\dfrac{1}{x^2+1}\,\mathrm{d}x =2x-2\arctan\left(x\right)$$

The integral is now solved and we yield the result :

$${\displaystyle\int}\ln\left(x^2+1\right)\,\mathrm{d}x=x\ln\left(x^2+1\right)+2\arctan\left(x\right)-2x + C$$

Answer 2

$2\int \frac{x^2}{x^2+1} $

Does not equal the expression with just an x in the numerator after the substitution . You can add subtract 1 in numerator to integrate $2\int \frac{x^2}{x^2+1} $

This is wrong :- $2\int \frac{x^2}{x^2+1} $ This is what you get after applying by-parts :- $2\int \frac{x^2}{x^2+1} dx $ Since differential of x is $dx$ . In by parts formula we have differential of $u $ not it's differenciation.