What is the integral of (x+2)/((x^2+4x+6)^5) ?

The integral of (x+2)/((x^2+4x+6)^5) is -1/(8(x^2+4x+6)^4)+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)/((x^2+4x+6)^5)

\int \frac{x + 2}{( x ^{2} + 4 x + 6 ) ^{5}} d x

- \frac{1}{8 ( x ^{2} + 4 x + 6 ) ^{4}} + C

Solution steps

\int \frac{x + 2}{( x ^{2} + 4 x + 6 ) ^{5}} d x

Apply u-substitution

: \int \frac{u}{( u ^{2} + 2 ) ^{5}} d u

= \int \frac{u}{( u ^{2} + 2 ) ^{5}} d u

Apply u-substitution

: \int \frac{1}{2 ( v + 2 ) ^{5}} d v

= \int \frac{1}{2 ( v + 2 ) ^{5}} 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 ) ^{5}} d v

Apply Integration By Parts

: \frac{v}{( v + 2 ) ^{5}} - \int - \frac{5 v}{( v + 2 ) ^{6}} d v

= \frac{1}{2} (\frac{v}{( v + 2 ) ^{5}} - \int - \frac{5 v}{( v + 2 ) ^{6}} d v)

\int - \frac{5 v}{( v + 2 ) ^{6}} d v = \frac{5}{4 ( v + 2 ) ^{4}} - \frac{2}{( v + 2 ) ^{5}}

= \frac{1}{2} (\frac{v}{( v + 2 ) ^{5}} - (\frac{5}{4 ( v + 2 ) ^{4}} - \frac{2}{( v + 2 ) ^{5}}))

Substitute back

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

Simplify

\frac{1}{2} \frac{( x + 2 ) ^{2}}{( ( x + 2 ) ^{2} + 2 ) ^{5}} - \frac{5}{4 ( ( x + 2 ) ^{2} + 2 ) ^{4}} - \frac{2}{( ( x + 2 ) ^{2} + 2 ) ^{5}} : - \frac{1}{8 ( x ^{2} + 4 x + 6 ) ^{4}}

= - \frac{1}{8 ( x ^{2} + 4 x + 6 ) ^{4}}

Add a constant to the solution

= - \frac{1}{8 ( x ^{2} + 4 x + 6 ) ^{4}} + C

\int 5 sin (x - 6) + 54 x^{8} + 9 sin (x - 6) d x

\int e^{2 x} cos (h) x d x

\int \frac{x ^{2} + 30 x - 9}{x ^{3} - 9 x} d x

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

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

What is the integral of (x+2)/((x^2+4x+6)^5) ?
The integral of (x+2)/((x^2+4x+6)^5) is -1/(8(x^2+4x+6)^4)+C