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

The integral of x^{-2}(4x^{-1}+3)^{3/4} is -((4+3x)^{7/4})/(7x^{7/4)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^{-2}(4x^{-1}+3)^{3/4}

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

- \frac{( 4 + 3 x ) ^{\frac{7}{4}}}{7 x ^{\frac{7}{4}}} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{4}{7} u^{\frac{7}{4}}

= - \frac{1}{4} \cdot \frac{4}{7} u^{\frac{7}{4}}

Substitute back

u = 4 x^{- 1} + 3

= - \frac{1}{4} \cdot \frac{4}{7} (4 x^{- 1} + 3)^{\frac{7}{4}}

Simplify

- \frac{1}{4} \cdot \frac{4}{7} (4 x^{- 1} + 3)^{\frac{7}{4}} : - \frac{( 4 + 3 x ) ^{\frac{7}{4}}}{7 x ^{\frac{7}{4}}}

= - \frac{( 4 + 3 x ) ^{\frac{7}{4}}}{7 x ^{\frac{7}{4}}}

Add a constant to the solution

= - \frac{( 4 + 3 x ) ^{\frac{7}{4}}}{7 x ^{\frac{7}{4}}} + C

1 0^{^{'}}

2 x y^{2} + x^{2} \frac{d y}{d x} = y^{2}

\int e^{- θ} cos (3 θ) d θ

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

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

What is the integral of x^{-2}(4x^{-1}+3)^{3/4} ?
The integral of x^{-2}(4x^{-1}+3)^{3/4} is -((4+3x)^{7/4})/(7x^{7/4)}+C