Can it be solved with arctan?

Question

Exercise:

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

The result to that exercise according to Symbolab

$$\ln \left|x\right|-\ln \left|x+1\right|+C$$

Can it be solved this way? As i did: $$\frac{1}{\sqrt{x}}\arctan\left(\sqrt{x}\right)+c$$

My solution is based on the following formula: $$\int \frac{du}{a^2+u^2}\:=\:\frac{1}{a}\arctan\left(\frac{u}{a}\right)+c$$

Answer 1

No, differentiating your answer yields: $$\frac{1}{2x\left(x+1\right)}-\dfrac{\arctan\left(\sqrt{x}\right)}{2x^\frac{3}{2}}$$ which obviously doesn't equal $\frac1{x+x^2}$.

Maybe you could show your workings?

Answer 2

Unfortunately, you result is wrong.

1. it simplifies to $\dfrac1{\sqrt x}\arctan\sqrt x+C$,

2. the derivative is $-\dfrac{\arctan\sqrt x}{2x\sqrt x}+\dfrac1{2x(x+1)}$.

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The correct solution is obtained by noticing

$$\frac1{x(x+1)}=\frac1x-\frac1{x+1}$$ which is immediate to integrate.

Answer 3

Hint: Use the fact that$$\frac1{x^2+x}=\frac1{x(x+1)}=\frac1x-\frac1{x+1}.$$

Answer 4

Hint

As an exercice you can try to use the formula with $arctan$ function ....note that $x^2+x=(x^2+x+ \frac 1 4)-\frac 1 4=(x+\frac 1 2)^2+(\frac i 2)^2$

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