What is the integral of 3x^{-4}(2+4x^{-3})^{-7} ?

The integral of 3x^{-4}(2+4x^{-3})^{-7} is (x^{18})/(24(2x^3+4)^6)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3x^{-4}(2+4x^{-3})^{-7}

\int 3 x^{- 4} (2 + 4 x^{- 3})^{- 7} d x

\frac{x ^{18}}{24 ( 2 x ^{3} + 4 ) ^{6}} + C

Solution steps

\int 3 x^{- 4} (2 + 4 x^{- 3})^{- 7} d x

Take the constant out:

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

= 3 \cdot \int x^{- 4} (2 + 4 x^{- 3})^{- 7} d x

Apply u-substitution

: \int - \frac{1}{12 u ^{7}} d u

= 3 \cdot \int - \frac{1}{12 u ^{7}} d u

Take the constant out:

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

= 3 (- \frac{1}{12} \cdot \int \frac{1}{u ^{7}} d u)

Apply the Power Rule

: - \frac{1}{6 u ^{6}}

= 3 (- \frac{1}{12} (- \frac{1}{6 u ^{6}}))

Substitute back

u = 2 + 4 x^{- 3}

= 3 (- \frac{1}{12} (- \frac{1}{6 ( 2 + 4 x ^{- 3} ) ^{6}}))

Simplify

3 (- \frac{1}{12} (- \frac{1}{6 ( 2 + 4 x ^{- 3} ) ^{6}})) : \frac{x ^{18}}{24 ( 2 x ^{3} + 4 ) ^{6}}

= \frac{x ^{18}}{24 ( 2 x ^{3} + 4 ) ^{6}}

Add a constant to the solution

= \frac{x ^{18}}{24 ( 2 x ^{3} + 4 ) ^{6}} + C

f^{^{'}} (t) - f (t) = - 4 t

\int_{- 8}^{1} \frac{1}{x ^{\frac{1}{3}}} d x

y^{^{'}} + 8 (tan (8 x)) y = - 2 cos (8 x)

\frac{( t ^{6} )}{( t ^{3} + 1 ) ^{2}}

\int (1 - sin (θ))^{2} d θ

What is the integral of 3x^{-4}(2+4x^{-3})^{-7} ?
The integral of 3x^{-4}(2+4x^{-3})^{-7} is (x^{18})/(24(2x^3+4)^6)+C