What is the integral of 1/(x(2x+5)^2) ?

The integral of 1/(x(2x+5)^2) is 1/25 ln|x|-1/25 ln|2x+5|+1/(5(2x+5))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x(2x+5)^2)

\int \frac{1}{x ( 2 x + 5 ) ^{2}} d x

\frac{1}{25} ln ∣ x ∣ - \frac{1}{25} ln ∣ 2 x + 5 ∣ + \frac{1}{5 ( 2 x + 5 )} + C

Solution steps

\int \frac{1}{x ( 2 x + 5 ) ^{2}} d x

Take the partial fraction of

\frac{1}{x ( 2 x + 5 ) ^{2}} : \frac{1}{25 x} - \frac{2}{25 ( 2 x + 5 )} - \frac{2}{5 ( 2 x + 5 ) ^{2}}

= \int \frac{1}{25 x} - \frac{2}{25 ( 2 x + 5 )} - \frac{2}{5 ( 2 x + 5 ) ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{25 x} d x - \int \frac{2}{25 ( 2 x + 5 )} d x - \int \frac{2}{5 ( 2 x + 5 ) ^{2}} d x

\int \frac{1}{25 x} d x = \frac{1}{25} ln ∣ x ∣

\int \frac{2}{25 ( 2 x + 5 )} d x = \frac{1}{25} ln ∣ 2 x + 5 ∣

\int \frac{2}{5 ( 2 x + 5 ) ^{2}} d x = - \frac{1}{5 ( 2 x + 5 )}

= \frac{1}{25} ln ∣ x ∣ - \frac{1}{25} ln ∣ 2 x + 5 ∣ - (- \frac{1}{5 ( 2 x + 5 )})

Simplify

= \frac{1}{25} ln ∣ x ∣ - \frac{1}{25} ln ∣ 2 x + 5 ∣ + \frac{1}{5 ( 2 x + 5 )}

Add a constant to the solution

= \frac{1}{25} ln ∣ x ∣ - \frac{1}{25} ln ∣ 2 x + 5 ∣ + \frac{1}{5 ( 2 x + 5 )} + C

\frac{d}{d x} (2 5^{x})

n = 0 \sum \infty 8 (- \frac{1}{2})^{n}

\frac{d}{d x} (8 x - 2)

\int 5 cos (x) - csc^{2} (x) d x

\int \frac{x}{81 x ^{4} - 1} d x

What is the integral of 1/(x(2x+5)^2) ?
The integral of 1/(x(2x+5)^2) is 1/25 ln|x|-1/25 ln|2x+5|+1/(5(2x+5))+C