What is the integral of 1/((x+3)^2(x-4)) ?

The integral of 1/((x+3)^2(x-4)) is -1/49 ln|x+3|+1/(7(x+3))+1/49 ln|x-4|+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+3)^2(x-4))

\int \frac{1}{( x + 3 ) ^{2} ( x - 4 )} d x

- \frac{1}{49} ln ∣ x + 3 ∣ + \frac{1}{7 ( x + 3 )} + \frac{1}{49} ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{1}{( x + 3 ) ^{2} ( x - 4 )} d x

Take the partial fraction of

\frac{1}{( x + 3 ) ^{2} ( x - 4 )} : - \frac{1}{49 ( x + 3 )} - \frac{1}{7 ( x + 3 ) ^{2}} + \frac{1}{49 ( x - 4 )}

= \int - \frac{1}{49 ( x + 3 )} - \frac{1}{7 ( x + 3 ) ^{2}} + \frac{1}{49 ( x - 4 )} 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}{49 ( x + 3 )} d x - \int \frac{1}{7 ( x + 3 ) ^{2}} d x + \int \frac{1}{49 ( x - 4 )} d x

\int \frac{1}{49 ( x + 3 )} d x = \frac{1}{49} ln ∣ x + 3 ∣

\int \frac{1}{7 ( x + 3 ) ^{2}} d x = - \frac{1}{7 ( x + 3 )}

\int \frac{1}{49 ( x - 4 )} d x = \frac{1}{49} ln ∣ x - 4 ∣

= - \frac{1}{49} ln ∣ x + 3 ∣ - (- \frac{1}{7 ( x + 3 )}) + \frac{1}{49} ln ∣ x - 4 ∣

Simplify

= - \frac{1}{49} ln ∣ x + 3 ∣ + \frac{1}{7 ( x + 3 )} + \frac{1}{49} ln ∣ x - 4 ∣

Add a constant to the solution

= - \frac{1}{49} ln ∣ x + 3 ∣ + \frac{1}{7 ( x + 3 )} + \frac{1}{49} ln ∣ x - 4 ∣ + C

\frac{d}{d x} (x + \frac{4}{x ^{2}})

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

\frac{\partial}{\partial x} ((x + 2 y)^{2})

f (x) = ln (x^{3})

\int ∣ 2 x - 1 ∣ d x

What is the integral of 1/((x+3)^2(x-4)) ?
The integral of 1/((x+3)^2(x-4)) is -1/49 ln|x+3|+1/(7(x+3))+1/49 ln|x-4|+C