What is the integral of (7x)/((1-4x)^4) ?

The integral of (7x)/((1-4x)^4) is (7(12x-1))/(96(-4x+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 (7x)/((1-4x)^4)

\int \frac{7 x}{( 1 - 4 x ) ^{4}} d x

\frac{7 ( 12 x - 1 )}{96 ( - 4 x + 1 ) ^{3}} + C

Solution steps

\int \frac{7 x}{( 1 - 4 x ) ^{4}} d x

Take the constant out:

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

= 7 \cdot \int \frac{x}{( 1 - 4 x ) ^{4}} d x

Apply u-substitution

: \int \frac{u - 1}{16 u ^{4}} d u

= 7 \cdot \int \frac{u - 1}{16 u ^{4}} d u

Take the constant out:

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

= 7 \cdot \frac{1}{16} \cdot \int \frac{u - 1}{u ^{4}} d u

Expand

\frac{u - 1}{u ^{4}} : \frac{1}{u ^{3}} - \frac{1}{u ^{4}}

= 7 \cdot \frac{1}{16} \cdot \int \frac{1}{u ^{3}} - \frac{1}{u ^{4}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 7 \cdot \frac{1}{16} (\int \frac{1}{u ^{3}} d u - \int \frac{1}{u ^{4}} d u)

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

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

= 7 \cdot \frac{1}{16} (- \frac{1}{2 u ^{2}} - (- \frac{1}{3 u ^{3}}))

Substitute back

u = 1 - 4 x

= 7 \cdot \frac{1}{16} (- \frac{1}{2 ( 1 - 4 x ) ^{2}} - (- \frac{1}{3 ( 1 - 4 x ) ^{3}}))

Simplify

7 \cdot \frac{1}{16} (- \frac{1}{2 ( 1 - 4 x ) ^{2}} - (- \frac{1}{3 ( 1 - 4 x ) ^{3}})) : \frac{7 ( 12 x - 1 )}{96 ( - 4 x + 1 ) ^{3}}

= \frac{7 ( 12 x - 1 )}{96 ( - 4 x + 1 ) ^{3}}

Add a constant to the solution

= \frac{7 ( 12 x - 1 )}{96 ( - 4 x + 1 ) ^{3}} + C

\int x^{2} \cdot sin (a x) d x

\frac{\partial}{\partial y} (e^{- 3 x} cos (3 y))

f (x) = ln (1 - x)

y^{^{'}} = - k y^{\frac{1}{2}}

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

What is the integral of (7x)/((1-4x)^4) ?
The integral of (7x)/((1-4x)^4) is (7(12x-1))/(96(-4x+1)^3)+C