Integral of (8x+4)/(4x^2+4x)

\int\frac{8x+4}{4x^2+4x}dx

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Answer

$\frac{1}{4}\int\frac{4}{x+x^2}dx+2\int\frac{x}{x+x^2}dx$

Step by step solution

Problem

$\int\frac{8x+4}{4x^2+4x}dx$
1

Taking the constant out of the integral

$\frac{1}{4}\int\frac{4+8x}{x+x^2}dx$
2

Split the fraction $\frac{8x+4}{x+x^2}$ in two terms with same denominator

$\frac{1}{4}\int\left(\frac{4}{x+x^2}+\frac{8x}{x+x^2}\right)dx$
3

The integral of a sum of two or more functions is equal to the sum of their integrals

$\frac{1}{4}\left(\int\frac{4}{x+x^2}dx+\int\frac{8x}{x+x^2}dx\right)$
4

Taking the constant out of the integral

$\frac{1}{4}\left(\int\frac{4}{x+x^2}dx+8\int\frac{x}{x+x^2}dx\right)$
5

Multiply $\left(8\int\frac{x}{x+x^2}dx+\int\frac{4}{x+x^2}dx\right)$ by $\frac{1}{4}$

$\frac{1}{4}\int\frac{4}{x+x^2}dx+2\int\frac{x}{x+x^2}dx$

Answer

$\frac{1}{4}\int\frac{4}{x+x^2}dx+2\int\frac{x}{x+x^2}dx$

Problem Analysis

Main topic:

Integral calculus

Time to solve it:

0.37 seconds

Views:

120