What is the integral of (x+2)/((x^2+4x+1)^2) ?

The integral of (x+2)/((x^2+4x+1)^2) is -1/(2(x^2+4x+1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x+2)/((x^2+4x+1)^2)

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

- \frac{1}{2 ( x ^{2} + 4 x + 1 )} + C

Solution steps

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

Apply u-substitution

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

Apply u-substitution

= \int \frac{1}{2 v ^{2}} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

= \frac{1}{2} (- \frac{1}{v})

Substitute back

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

Simplify

\frac{1}{2} (- \frac{1}{( x + 2 ) ^{2} - 3}) : - \frac{1}{2 ( x ^{2} + 4 x + 1 )}

= - \frac{1}{2 ( x ^{2} + 4 x + 1 )}

Add a constant to the solution

= - \frac{1}{2 ( x ^{2} + 4 x + 1 )} + C

What is the integral of (x+2)/((x^2+4x+1)^2) ?
The integral of (x+2)/((x^2+4x+1)^2) is -1/(2(x^2+4x+1))+C