What is the integral of (x^2+6x)/((x+3)^2) ?

The integral of (x^2+6x)/((x+3)^2) is -(x^2+6x)/(x+3)+2x+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+6x)/((x+3)^2)

\int \frac{x ^{2} + 6 x}{( x + 3 ) ^{2}} d x

- \frac{x ^{2} + 6 x}{x + 3} + 2 x + C

Solution steps

\int \frac{x ^{2} + 6 x}{( x + 3 ) ^{2}} d x

Apply Integration By Parts

: - \frac{x ^{2} + 6 x}{x + 3} - \int - 2 d x

= - \frac{x ^{2} + 6 x}{x + 3} - \int - 2 d x

\int - 2 d x = - 2 x

= - \frac{x ^{2} + 6 x}{x + 3} - (- 2 x)

Simplify

= - \frac{x ^{2} + 6 x}{x + 3} + 2 x

Add a constant to the solution

= - \frac{x ^{2} + 6 x}{x + 3} + 2 x + C

\int \frac{0}{x + 2} d x

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

\int (5 x - 6) d x

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

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

What is the integral of (x^2+6x)/((x+3)^2) ?
The integral of (x^2+6x)/((x+3)^2) is -(x^2+6x)/(x+3)+2x+C