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

The integral of x(x^2-1)^{99} is 1/200 (x^2-1)^{100}+C

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

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

\frac{1}{200} (x^{2} - 1)^{100} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{99}}{2} d u

= \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

= \frac{1}{2} \cdot \int u^{99} d u

Apply the Power Rule

: \frac{u ^{100}}{100}

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

Substitute back

u = x^{2} - 1

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

Simplify

\frac{1}{2} \cdot \frac{( x ^{2} - 1 ) ^{100}}{100} : \frac{1}{200} (x^{2} - 1)^{100}

= \frac{1}{200} (x^{2} - 1)^{100}

Add a constant to the solution

= \frac{1}{200} (x^{2} - 1)^{100} + C

n = 0 \sum \infty \frac{1}{n + 9}

\int x^{3} + 3 d x

\frac{d y}{d x} = 2 x - y, y (0) = 1

x \to 2 lim (x^{2} + 3 x + 5)

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

What is the integral of x(x^2-1)^{99} ?
The integral of x(x^2-1)^{99} is 1/200 (x^2-1)^{100}+C