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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

u = x + 1

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

Add a constant to the solution

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

\int sin^{3} (5 x) d x

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

\int_{0}^{20} 4 + 3 x d x

y^{^{''}} - 3 y^{^{'}} - 4 y = - 3, y (0) = 0, y^{^{'}} (0) = 0

\int tan^{2} (2 x se) c^{2} 2 x

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