What is the integral of x/((x-3)^3) ?

The integral of x/((x-3)^3) is -1/(x-3)-3/(2(x-3)^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x/((x-3)^3)

\int \frac{x}{( x - 3 ) ^{3}} d x

- \frac{1}{x - 3} - \frac{3}{2 ( x - 3 ) ^{2}} + C

Solution steps

\int \frac{x}{( x - 3 ) ^{3}} d x

Apply u-substitution

: \int \frac{u + 3}{u ^{3}} d u

= \int \frac{u + 3}{u ^{3}} d u

Expand

\frac{u + 3}{u ^{3}} : \frac{1}{u ^{2}} + \frac{3}{u ^{3}}

= \int \frac{1}{u ^{2}} + \frac{3}{u ^{3}} 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

= \int \frac{1}{u ^{2}} d u + \int \frac{3}{u ^{3}} d u

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

\int \frac{3}{u ^{3}} d u = - \frac{3}{2 u ^{2}}

= - \frac{1}{u} - \frac{3}{2 u ^{2}}

Substitute back

u = x - 3

= - \frac{1}{x - 3} - \frac{3}{2 ( x - 3 ) ^{2}}

Add a constant to the solution

= - \frac{1}{x - 3} - \frac{3}{2 ( x - 3 ) ^{2}} + C

\int \frac{1}{( 4 x ^{2} - 4 x + 5 ) ^{2}} d x

\int sin (2 x) x^{3} d x

\int \frac{1}{9} csc^{2} (x) + \frac{1}{x ^{2}} d x

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

\int \frac{1}{y 9 y ^{2} - 16} d y

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