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

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

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

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

Solution steps

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

Expand

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

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

\int \frac{x ^{2}}{( x - 2 ) ^{3}} d x = - \frac{x ^{2}}{2 ( x - 2 ) ^{2}} + ln ∣ x - 2 ∣ - \frac{2}{x - 2}

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

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

Add a constant to the solution

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

\int \frac{x + 2}{x ^{2}} d x

2^{x}, - 4, 4

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

\frac{d}{d x} (\frac{ln ( x )}{sin ( x )})

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

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