# Finding the improper integral $\int^\infty_0\frac{1}{x^3+1}\,dx$ [duplicate]

$$\int^\infty_0\frac{1}{x^3+1}\,dx$$

The answer is $\frac{2\pi}{3\sqrt{3}}$.

How can I evaluate this integral?

$$\frac1{1+x^3}=\frac A{1+x}+\frac{Bx+C}{1-x+x^2}$$

Multiply either sides by $(1+x)(1-x+x^2)$ and compare the coefficients of the different powers of $x$ to find $A,B,C$

As $\displaystyle1-x+x^2=\frac{4x^2-4x+4}4=\frac{(2x-1)^2+3}4$

Using Trigonometric substitution, set $\displaystyle 2x-1=\sqrt3\tan\phi$ in the second part

• thank you this was the part I couldn't solve :) – Pumpkin Jan 3 '14 at 12:34
• @Pumpkin, hope you can find $B,C$ with the hint – lab bhattacharjee Jan 3 '14 at 12:35
• I have found them before but I couldnt do that $\displaystyle 2x-1=\sqrt3\tan\phi$ thanks for the solution – Pumpkin Jan 3 '14 at 12:36
• @Pumpkin, tome , Trigonometric Identities seem indispensable for integration – lab bhattacharjee Jan 3 '14 at 12:39

In general, when you have $$\frac{Bx+C}{ax^2+bx+c} = \frac{Bx}{ax^2+bx+c} + \frac{C}{ax^2+bx+c}$$

Then the left addend can be easily integrated by multiplying by $2a/B$ (and dividing outside the integral sign). For the left addend, assuming we can't factorize it in $\mathbb R$, the procedure I'd use is to write the denominator $ax^2 + bx + c$ as $(x+\alpha)^2 + \beta$ (first make it monic and then complete the square), where $\alpha$ and $\beta$ are real numbers.

This can also be written as $\beta\left((\frac{x+\alpha}{\sqrt \beta})^2+1\right)$ so you end up with $$\frac{\tilde C}{\left(\frac{x+\alpha}{\sqrt \beta}\right)^2+1}$$

($\tilde C$ is the real number resulting from the previous steps) which is the derivative of $$\arctan{\left(\frac{x+\alpha}{\sqrt\beta}\right)} + k$$

$$x^3+1 = (x+1)(x^2-x+1)$$ Logic: Do partial fraction decomposition.Find $A,B,C$.

$$\frac{1}{x^3+1} = \frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}$$ By comparing corresponding co-efficients of different powers of $x$, you will end up with equations in A,B,C.After solving you get : $$A=\frac{1}{3},B=\frac{-1}{3},C=\frac{2}{3}$$ Then use this: $$\int\frac{1}{x}\,dx=\log x+c$$

• I did partial fraction but I couldnt figure out how to solve second part $\frac{Bx+C}{x^2-x+1}$ – Pumpkin Jan 3 '14 at 12:32
• $\int\frac{1}{x}\,dx=\log x+c$ cannot be true because base of nonspecified log is 10? – Pumpkin Jan 3 '14 at 15:00