What is the integral of 2x(x^2-1)^{97} ?

The integral of 2x(x^2-1)^{97} is 1/98 (x^2-1)^{98}+C

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integral of 2x(x^2-1)^{97}

\int 2 x (x^{2} - 1)^{97} d x

\frac{1}{98} (x^{2} - 1)^{98} + C

Solution steps

\int 2 x (x^{2} - 1)^{97} 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)^{97} d x

Apply u-substitution

: \int \frac{u ^{97}}{2} d u

= 2 \cdot \int \frac{u ^{97}}{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^{97} d u

Apply the Power Rule

: \frac{u ^{98}}{98}

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{98}}{98}

Substitute back

u = x^{2} - 1

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 1 ) ^{98}}{98}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 1 ) ^{98}}{98} : \frac{1}{98} (x^{2} - 1)^{98}

= \frac{1}{98} (x^{2} - 1)^{98}

Add a constant to the solution

= \frac{1}{98} (x^{2} - 1)^{98} + C

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What is the integral of 2x(x^2-1)^{97} ?
The integral of 2x(x^2-1)^{97} is 1/98 (x^2-1)^{98}+C