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

Question

I need to integrate $\frac{x}{1+x^2}$ and I tried it like this: $$f(x)=\frac{x}{1+x^2}=\frac{x}{(x+1)(x-1)}=\frac{x+1-1}{(x+1)(x-1)}= \frac{x+1}{(x+1)(x-1)}-\frac{1}{(x+1)(x-1)}=\frac{1}{x-1}-\frac{1}{1+x^2}$$ After I simplified it I integrate it! $$\int{f(x)}=\int{\frac{1}{x-1}}-\int{\frac{1}{1+x^2}}$$

$\int{\frac{dx}{1+x^2}}$ becomes arctan(x) and I substitute $\int{\frac{dx}{x-1}}$ to $\int{\frac{du}{u}}$ now it becomes $ln(x-1)$ after this been done I get: $$\int{f(x)}=ln(x-1)-arctan(x)$$ but when I type integrate x/(1+x^2) into wolframAlpha I it tells me the answer is: $$\int{f(x)}=\frac{1}{2}log(x^2+1)$$ where did I go wrong or what should I have done instead ?

Thank you for your help in advance,

Raavgo

Answer 1

The mistake is at the very beginning: it is not true that $1+x^2 = (1-x)(1+x)$. What you thought about is $1\color{red} - x^2 = (1-x)(1+x)$. (Also, remember that the roots of $1+x^2$ are $\pm \Bbb i$, therefore your factorization is clearly wrong.)

The exercise is much simpler if you notice that $x = \frac 1 2 (1+x^2)'$, therefore

$$\int \frac x {x^2 + 1} \ \Bbb d x = \int \frac {\frac 1 2 (x^2 + 1)'} {x^2 + 1} \ \Bbb d x = \frac 1 2 \log |x^2 + 1| + C ,$$

where $C \in \Bbb R$ is an integration constant.

(I have used the well-known formula that $\int \frac {f'} f \ \Bbb d x = \log |f| + C$.)

Answer 2

Hint:

Use the substitution $$1+x^2=t \quad \rightarrow \quad 2xdx=dt$$

Answer 3

Easiest way is to use $$\int \frac{f'(x)}{f(x)} dx = \log(f(x)) + c$$

Answer 4

I would integrate 'by inspection', using the standard result stated above:

$$ \int \frac{f'(x) } {f(x) } \quad dx= ln(f(x)) +c$$

Therefore, answering your question;

$$\int \frac{x} {1+x^2} \quad dx, $$

If $f(x) =x^2 \, then, \, \, f'(x) =2x$

Another equivalent form of the integral becomes;

$${1\over2} \cdot \int \frac{2x} {1+x^2} \quad dx $$

The integral is in the form of $\frac{f' (x) } {f(x) }, $

So the integral may be evaluated as;

$${1 \over 2} \cdot ln(1+x^2)+c$$

Hope this helps.