What is the integral of 2x(x^2-1)^{99} ?

The integral of 2x(x^2-1)^{99} is 1/100 (x^2-1)^{100}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 2x(x^2-1)^{99}

\int 2 x (x^{2} - 1)^{99} d x

\frac{1}{100} (x^{2} - 1)^{100} + C

Solution steps

\int 2 x (x^{2} - 1)^{99} 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)^{99} d x

Apply u-substitution

= 2 \cdot \int \frac{u ^{99}}{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^{99} d u

Apply the Power Rule

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{100}}{100}

Substitute back

u = x^{2} - 1

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 1 ) ^{100}}{100}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 1 ) ^{100}}{100} : \frac{1}{100} (x^{2} - 1)^{100}

= \frac{1}{100} (x^{2} - 1)^{100}

Add a constant to the solution

= \frac{1}{100} (x^{2} - 1)^{100} + C

What is the integral of 2x(x^2-1)^{99} ?
The integral of 2x(x^2-1)^{99} is 1/100 (x^2-1)^{100}+C