What is the integral of 200x(x^2-1)^{99} ?

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

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integral of 200x(x^2-1)^{99}

\int 200 x (x^{2} - 1)^{99} d x

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

Solution steps

\int 200 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

= 200 \cdot \int x (x^{2} - 1)^{99} d x

Apply u-substitution

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

= 200 \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

= 200 \cdot \frac{1}{2} \cdot \int u^{99} d u

Apply the Power Rule

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

= 200 \cdot \frac{1}{2} \cdot \frac{u ^{100}}{100}

Substitute back

u = x^{2} - 1

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

Simplify

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

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

Add a constant to the solution

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

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