What is the integral of 1/((x^2-49)^2) ?

The integral of 1/((x^2-49)^2) is 1/1372 ln|x+7|-1/(196(x+7))-1/1372 ln|x-7|-1/(196(x-7))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/((x^2-49)^2)

\int \frac{1}{( x ^{2} - 49 ) ^{2}} d x

\frac{1}{1372} ln ∣ x + 7 ∣ - \frac{1}{196 ( x + 7 )} - \frac{1}{1372} ln ∣ x - 7 ∣ - \frac{1}{196 ( x - 7 )} + C

Solution steps

\int \frac{1}{( x ^{2} - 49 ) ^{2}} d x

Take the partial fraction of

\frac{1}{( x ^{2} - 49 ) ^{2}} : \frac{1}{1372 ( x + 7 )} + \frac{1}{196 ( x + 7 ) ^{2}} - \frac{1}{1372 ( x - 7 )} + \frac{1}{196 ( x - 7 ) ^{2}}

= \int \frac{1}{1372 ( x + 7 )} + \frac{1}{196 ( x + 7 ) ^{2}} - \frac{1}{1372 ( x - 7 )} + \frac{1}{196 ( x - 7 ) ^{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}{1372 ( x + 7 )} d x + \int \frac{1}{196 ( x + 7 ) ^{2}} d x - \int \frac{1}{1372 ( x - 7 )} d x + \int \frac{1}{196 ( x - 7 ) ^{2}} d x

\int \frac{1}{1372 ( x + 7 )} d x = \frac{1}{1372} ln ∣ x + 7 ∣

\int \frac{1}{196 ( x + 7 ) ^{2}} d x = - \frac{1}{196 ( x + 7 )}

\int \frac{1}{1372 ( x - 7 )} d x = \frac{1}{1372} ln ∣ x - 7 ∣

\int \frac{1}{196 ( x - 7 ) ^{2}} d x = - \frac{1}{196 ( x - 7 )}

= \frac{1}{1372} ln ∣ x + 7 ∣ - \frac{1}{196 ( x + 7 )} - \frac{1}{1372} ln ∣ x - 7 ∣ - \frac{1}{196 ( x - 7 )}

Add a constant to the solution

= \frac{1}{1372} ln ∣ x + 7 ∣ - \frac{1}{196 ( x + 7 )} - \frac{1}{1372} ln ∣ x - 7 ∣ - \frac{1}{196 ( x - 7 )} + C

What is the integral of 1/((x^2-49)^2) ?
The integral of 1/((x^2-49)^2) is 1/1372 ln|x+7|-1/(196(x+7))-1/1372 ln|x-7|-1/(196(x-7))+C