Integral of (x^2-4)/x

\int\frac{x^2-4}{x}dx

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Answer

$-4\ln\left|x\right|+\frac{1}{2}x^2+C_0$

Step by step solution

Problem

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

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

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

Simplifying the fraction by $x$

$\int\left(\frac{-4}{x}+x\right)dx$
3

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

$\int\frac{-4}{x}dx+\int xdx$
4

Applying the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$\int\frac{-4}{x}dx+\frac{1}{2}x^2$
5

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$\frac{1}{2}x^2-4\ln\left|x\right|$
6

Add the constant of integration

$-4\ln\left|x\right|+\frac{1}{2}x^2+C_0$

Answer

$-4\ln\left|x\right|+\frac{1}{2}x^2+C_0$

Problem Analysis

Main topic:

Integral calculus

Time to solve it:

0.24 seconds

Views:

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