What is the integral from 0 to x of (6x)/((1+x)^4) ?

The integral from 0 to x of (6x)/((1+x)^4) is 6(-1/(2(x+1)^2)+1/(3(x+1)^3)+1/2-1/3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to x of (6x)/((1+x)^4)

\int_{0}^{x} \frac{6 x}{( 1 + x ) ^{4}} d x

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

Solution steps

\int_{0}^{x} \frac{6 x}{( 1 + x ) ^{4}} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{x} \frac{x}{( 1 + x ) ^{4}} d x

Apply u-substitution

: \int_{1}^{1 + x} \frac{u - 1}{u ^{4}} d u

= 6 \cdot \int_{1}^{1 + x} \frac{u - 1}{u ^{4}} d u

Expand

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

= 6 \cdot \int_{1}^{1 + x} \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

= 6 (\int_{1}^{1 + x} \frac{1}{u ^{3}} d u - \int_{1}^{1 + x} \frac{1}{u ^{4}} d u)

\int_{1}^{1 + x} \frac{1}{u ^{3}} d u = - \frac{1}{2 ( 1 + x ) ^{2}} + \frac{1}{2}

\int_{1}^{1 + x} \frac{1}{u ^{4}} d u = - \frac{1}{3 ( 1 + x ) ^{3}} + \frac{1}{3}

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

Simplify

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

\int_{0}^{1} 2^{x} + x d x

\int_{0}^{1} \frac{1}{x ^{4} + 1} d x

\int_{- 2}^{2} (x^{2}) d x

\int_{0}^{4} (x^{2} - 2 y^{2}) d x

\int_{1}^{5} (2 x + \frac{1}{x}) d x

What is the integral from 0 to x of (6x)/((1+x)^4) ?
The integral from 0 to x of (6x)/((1+x)^4) is 6(-1/(2(x+1)^2)+1/(3(x+1)^3)+1/2-1/3)