What is the integral of (6x+5)/((6x^2+10x)^5) ?

The integral of (6x+5)/((6x^2+10x)^5) is -1/(8(6x^2+10x)^4)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (6x+5)/((6x^2+10x)^5)

\int \frac{6 x + 5}{( 6 x ^{2} + 10 x ) ^{5}} d x

- \frac{1}{8 ( 6 x ^{2} + 10 x ) ^{4}} + C

Solution steps

\int \frac{6 x + 5}{( 6 x ^{2} + 10 x ) ^{5}} d x

Apply u-substitution

: \int \frac{1}{2 u ^{5}} d u

= \int \frac{1}{2 u ^{5}} d u

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}{u ^{5}} d u

Apply the Power Rule

: - \frac{1}{4 u ^{4}}

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

Substitute back

u = 6 x^{2} + 10 x

= \frac{1}{2} (- \frac{1}{4 ( 6 x ^{2} + 10 x ) ^{4}})

Simplify

\frac{1}{2} (- \frac{1}{4 ( 6 x ^{2} + 10 x ) ^{4}}) : - \frac{1}{8 ( 6 x ^{2} + 10 x ) ^{4}}

= - \frac{1}{8 ( 6 x ^{2} + 10 x ) ^{4}}

Add a constant to the solution

= - \frac{1}{8 ( 6 x ^{2} + 10 x ) ^{4}} + C

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

x \to 0 lim (- 3)

\frac{d}{d t} (e^{\frac{t}{7}} sin (\frac{t}{7}))

\frac{\partial}{\partial x} (y + \frac{2}{x ^{2}})

7 x^{2} + x y + 7 y^{2} = 15, (1, 1)

What is the integral of (6x+5)/((6x^2+10x)^5) ?
The integral of (6x+5)/((6x^2+10x)^5) is -1/(8(6x^2+10x)^4)+C