why integration of $y=\frac{1}{1+x^2}$ is $\arctan(x)+c$?

Question

I manually integrated $$\int \frac {dx}{x^2+1}$$ using u substitution and I got $$\frac {\ln(1+x^2)}{2x}$$ where did I go wrong?

u substitution

Answer 1

You can't take the $\frac{1}{2x}$ out of the integral. That only works with constants.

Answer 2

$x$ depends on $u$ hence you have to express $x$ as a function of $u$.

However, if $g=f^{-1}$ then $$ g'(y)=\frac{1}{f'\left(g\left(y\right)\right)} $$ With $f=\tan$ then $$ \text{arctan}'\left(y\right)=\frac{1}{\tan'\left(\text{arctan}\left(y\right)\right)} $$ Using that $\tan'\left(x\right)=1+\tan^2(x)$ we have $$ \text{arctan}'\left(y\right)=\frac{1}{1+\tan\left(\text{arctan}\left(y\right)\right)}=\frac{1}{1+y^2} $$

Answer 3

As you know an indefinite integral such as $\int \frac {dx}{1+x^2}$ is an antiderivative.

Since $$ \frac {d}{dx} \tan ^{-1}x = \frac {1}{1+x^2}$$ we have $$\int \frac {dx}{1+x^2}= tan ^{-1}x +C$$

Back to your work,

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

$$\int \frac {dx}{1+x^2}= \int \frac {du}{u(2x)}=??? $$

What did you do with the $(2x)$ ?

Answer 4

You can't put outside the integral the $1/2x$ $$I=\int \frac {dx}{1+x^2}$$ $$ \text {for this substitution } u=x^2+1 \implies du=2xdx$$ Then $$I=\int \frac {dx}{1+x^2}=\int \frac {du}{2xu}$$ You cant go further...

Answer 5

In your case, after u-substitution, then$$\int du\,\frac {2x}u\neq\frac {\log u}{2x}$$

Answer 6

Note that you wrote down $u=1+x^2$ in the first place, so you can't just pull $\frac{1}{2x}$ out.

Sometimes it's hard to evaluate integrals. Therefore, there have been ways and methods to solve them. I think you need trigonometric substitution. See Tri-sub on Wiki. Just replace $a$ with $1$.

Answer 7

You can get the result indirectly through the chain rule, as follows:

Let y = arctanx. Then tany = x. Differentiate w/r to x: you get dy/dx = the square of cosy. But cosy = 1/[sq. root of(1+x-squared]) ....[To see this, draw a right triangle with legs 1 and x as the adjacent sides of angle y (to conform to the condition that tany=x)]. Thus we have dy/dx=1/(1+x-squared), i.e. d(arctanx)/dx=1/(1+x-squared), hence arctanx = the integral (antiderivative) of 1/(1+x-squared).