What is the integral of (2x+100)^{-3} ?

The integral of (2x+100)^{-3} is -1/(4(2x+100)^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x+100)^{-3}

\int (2 x + 100)^{- 3} d x

- \frac{1}{4 ( 2 x + 100 ) ^{2}} + C

Solution steps

\int (2 x + 100)^{- 3} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \int \frac{1}{( 2 x + 100 ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{2 u ^{3}} d u

= \int \frac{1}{2 u ^{3}} 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 ^{3}} d u

Apply the Power Rule

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

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

Substitute back

u = 2 x + 100

= \frac{1}{2} (- \frac{1}{2 ( 2 x + 100 ) ^{2}})

Simplify

\frac{1}{2} (- \frac{1}{2 ( 2 x + 100 ) ^{2}}) : - \frac{1}{4 ( 2 x + 100 ) ^{2}}

= - \frac{1}{4 ( 2 x + 100 ) ^{2}}

Add a constant to the solution

= - \frac{1}{4 ( 2 x + 100 ) ^{2}} + C

\int e y^{2} d y

\int x^{\frac{1}{4}} + 9 d x

\int sec (x) (9 sec (x) + 7 tan (x)) d x

\int \frac{x + 1}{1 + x} d x

\int (tan (x))^{2} (sec (x))^{2} d x

What is the integral of (2x+100)^{-3} ?
The integral of (2x+100)^{-3} is -1/(4(2x+100)^2)+C