What is the integral of 2x(x^2+1)^7 ?

The integral of 2x(x^2+1)^7 is 1/8 (x^2+1)^8+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2x(x^2+1)^7

\int 2 x (x^{2} + 1)^{7} d x

\frac{1}{8} (x^{2} + 1)^{8} + C

Solution steps

\int 2 x (x^{2} + 1)^{7} d x

Take the constant out:

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

= 2 \cdot \int x (x^{2} + 1)^{7} d x

Apply u-substitution

: \int \frac{u ^{7}}{2} d u

= 2 \cdot \int \frac{u ^{7}}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int u^{7} d u

Apply the Power Rule

: \frac{u ^{8}}{8}

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{8}}{8}

Substitute back

u = x^{2} + 1

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 1 ) ^{8}}{8}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 1 ) ^{8}}{8} : \frac{1}{8} (x^{2} + 1)^{8}

= \frac{1}{8} (x^{2} + 1)^{8}

Add a constant to the solution

= \frac{1}{8} (x^{2} + 1)^{8} + C

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

\int \frac{1}{4 - v ^{2}} d v

\int \frac{3}{x + 1} d x

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

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

What is the integral of 2x(x^2+1)^7 ?
The integral of 2x(x^2+1)^7 is 1/8 (x^2+1)^8+C