What is the integral of (3x^2+1)^36x ?

The integral of (3x^2+1)^36x is 1/4 (3x^2+1)^4+C

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integral of (3x^2+1)^36x

\int (3 x^{2} + 1)^{3} 6 x d x

\frac{1}{4} (3 x^{2} + 1)^{4} + C

Solution steps

\int (3 x^{2} + 1)^{3} \cdot 6 x d x

Take the constant out:

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

= 6 \cdot \int (3 x^{2} + 1)^{3} x d x

Apply u-substitution

= 6 \cdot \int \frac{u ^{3}}{6} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int u^{3} d u

Apply the Power Rule

= 6 \cdot \frac{1}{6} \cdot \frac{u ^{4}}{4}

Substitute back

u = 3 x^{2} + 1

= 6 \cdot \frac{1}{6} \cdot \frac{( 3 x ^{2} + 1 ) ^{4}}{4}

Simplify

6 \cdot \frac{1}{6} \cdot \frac{( 3 x ^{2} + 1 ) ^{4}}{4} : \frac{1}{4} (3 x^{2} + 1)^{4}

= \frac{1}{4} (3 x^{2} + 1)^{4}

Add a constant to the solution

= \frac{1}{4} (3 x^{2} + 1)^{4} + C

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What is the integral of (3x^2+1)^36x ?
The integral of (3x^2+1)^36x is 1/4 (3x^2+1)^4+C