What is the integral of (2x^2+3)^4x ?

The integral of (2x^2+3)^4x is 1/20 (2x^2+3)^5+C

integral of (2x^2+3)^4x

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

\frac{1}{20} (2 x^{2} + 3)^{5} + C

Solution steps

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

Apply u-substitution

= \int \frac{u ^{4}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u^{4} d u

Apply the Power Rule

= \frac{1}{4} \cdot \frac{u ^{5}}{5}

Substitute back

u = 2 x^{2} + 3

= \frac{1}{4} \cdot \frac{( 2 x ^{2} + 3 ) ^{5}}{5}

Simplify

\frac{1}{4} \cdot \frac{( 2 x ^{2} + 3 ) ^{5}}{5} : \frac{1}{20} (2 x^{2} + 3)^{5}

= \frac{1}{20} (2 x^{2} + 3)^{5}

Add a constant to the solution

= \frac{1}{20} (2 x^{2} + 3)^{5} + C

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What is the integral of (2x^2+3)^4x ?
The integral of (2x^2+3)^4x is 1/20 (2x^2+3)^5+C