What is the integral of 1/((3x+5)^4) ?

The integral of 1/((3x+5)^4) is -1/(9(3x+5)^3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((3x+5)^4)

\int \frac{1}{( 3 x + 5 ) ^{4}} d x

- \frac{1}{9 ( 3 x + 5 ) ^{3}} + C

Solution steps

\int \frac{1}{( 3 x + 5 ) ^{4}} d x

Apply u-substitution

: \int \frac{1}{3 u ^{4}} d u

= \int \frac{1}{3 u ^{4}} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= \frac{1}{3} (- \frac{1}{3 u ^{3}})

Substitute back

u = 3 x + 5

= \frac{1}{3} (- \frac{1}{3 ( 3 x + 5 ) ^{3}})

Simplify

\frac{1}{3} (- \frac{1}{3 ( 3 x + 5 ) ^{3}}) : - \frac{1}{9 ( 3 x + 5 ) ^{3}}

= - \frac{1}{9 ( 3 x + 5 ) ^{3}}

Add a constant to the solution

= - \frac{1}{9 ( 3 x + 5 ) ^{3}} + C

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

y^{^{'}} = 3 y + 5

\int - 10 sin (x) d x

What is the integral of 1/((3x+5)^4) ?
The integral of 1/((3x+5)^4) is -1/(9(3x+5)^3)+C