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

The integral of 1/((x^2-2)^2) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

8 (\frac{1}{64 2} ln 2 x + 22 - \frac{1}{32 ( 2 x + 2 2 )} - \frac{1}{64 2} ln 2 x - 22 - \frac{1}{32 ( 2 x - 2 2 )}) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{8}{( u ^{2} - 8 ) ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int \frac{1}{( u ^{2} - 8 ) ^{2}} d u

Take the partial fraction of

\frac{1}{( u ^{2} - 8 ) ^{2}} : \frac{1}{64 2 ( u + 2 2 )} + \frac{1}{32 ( u + 2 2 ) ^{2}} - \frac{1}{64 2 ( u - 2 2 )} + \frac{1}{32 ( u - 2 2 ) ^{2}}

= 8 \cdot \int \frac{1}{64 2 ( u + 2 2 )} + \frac{1}{32 ( u + 2 2 ) ^{2}} - \frac{1}{64 2 ( u - 2 2 )} + \frac{1}{32 ( u - 2 2 ) ^{2}} d u

Apply the Sum Rule:

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

= 8 (\int \frac{1}{64 2 ( u + 2 2 )} d u + \int \frac{1}{32 ( u + 2 2 ) ^{2}} d u - \int \frac{1}{64 2 ( u - 2 2 )} d u + \int \frac{1}{32 ( u - 2 2 ) ^{2}} d u)

\int \frac{1}{64 2 ( u + 2 2 )} d u = \frac{1}{64 2} ln u + 22

\int \frac{1}{32 ( u + 2 2 ) ^{2}} d u = - \frac{1}{32 ( u + 2 2 )}

\int \frac{1}{64 2 ( u - 2 2 )} d u = \frac{1}{64 2} ln u - 22

\int \frac{1}{32 ( u - 2 2 ) ^{2}} d u = - \frac{1}{32 ( u - 2 2 )}

= 8 (\frac{1}{64 2} ln u + 22 - \frac{1}{32 ( u + 2 2 )} - \frac{1}{64 2} ln u - 22 - \frac{1}{32 ( u - 2 2 )})

Substitute back

u = 2 x

= 8 (\frac{1}{64 2} ln 2 x + 22 - \frac{1}{32 ( 2 x + 2 2 )} - \frac{1}{64 2} ln 2 x - 22 - \frac{1}{32 ( 2 x - 2 2 )})

Add a constant to the solution

= 8 (\frac{1}{64 2} ln 2 x + 22 - \frac{1}{32 ( 2 x + 2 2 )} - \frac{1}{64 2} ln 2 x - 22 - \frac{1}{32 ( 2 x - 2 2 )}) + C

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

x^{^{'}} = 2 (x + 1)

\frac{d}{d x} (7 x^{2} sin (x) tan (x))

y = x + x, (4, 6)

\int \frac{1}{tan ( v )} d v