What is the integral of 1/((1+x+y+z)^3) ?

The integral of 1/((1+x+y+z)^3) is -1/(2(1+x+y+z)^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((1+x+y+z)^3)

\int \frac{1}{( 1 + x + y + z ) ^{3}} d z

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

Solution steps

\int \frac{1}{( 1 + x + y + z ) ^{3}} d z

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= - \frac{1}{2 u ^{2}}

Substitute back

u = 1 + x + y + z

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

Add a constant to the solution

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

\int θ tan (6) θ^{2} d θ

\int e^{- s t} \cdot t \cdot cos (t) d t

\int \frac{3 x ^{2} - 8 x}{x ^{3} - 4 x ^{2} + 1} d x

\int x e^{\frac{- x ^{2}}{a}} d x

\int y \cdot ln (y) d y

What is the integral of 1/((1+x+y+z)^3) ?
The integral of 1/((1+x+y+z)^3) is -1/(2(1+x+y+z)^2)+C