What is the integral of (3x^2-1)^43x ?

The integral of (3x^2-1)^43x is 1/10 (3x^2-1)^5+C

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integral of (3x^2-1)^43x

\int (3 x^{2} - 1)^{4} 3 x d x

\frac{1}{10} (3 x^{2} - 1)^{5} + C

Solution steps

\int (3 x^{2} - 1)^{4} \cdot 3 x d x

Take the constant out:

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

= 3 \cdot \int (3 x^{2} - 1)^{4} x d x

Apply u-substitution

: \int \frac{( 3 u - 1 ) ^{4}}{2} d u

= 3 \cdot \int \frac{( 3 u - 1 ) ^{4}}{2} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int (3 u - 1)^{4} d u

Apply u-substitution

: \int \frac{v ^{4}}{3} d v

= 3 \cdot \frac{1}{2} \cdot \int \frac{v ^{4}}{3} d v

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \int v^{4} d v

Apply the Power Rule

: \frac{v ^{5}}{5}

= 3 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{v ^{5}}{5}

Substitute back

= 3 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{( 3 x ^{2} - 1 ) ^{5}}{5}

Simplify

3 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{( 3 x ^{2} - 1 ) ^{5}}{5} : \frac{1}{10} (3 x^{2} - 1)^{5}

= \frac{1}{10} (3 x^{2} - 1)^{5}

Add a constant to the solution

= \frac{1}{10} (3 x^{2} - 1)^{5} + C

\int (x^{3} + 5) d x

x^{2} - 6 (3) + 9

3 x^{2} + 2

\int \frac{x}{4 x ^{2} + 1} d x

\int e^{\frac{t}{5}} d t

What is the integral of (3x^2-1)^43x ?
The integral of (3x^2-1)^43x is 1/10 (3x^2-1)^5+C