What is the integral of (x^2+3x-1)^4(2x+3) ?

The integral of (x^2+3x-1)^4(2x+3) is 1/5120 (4x^2+12x-4)^5+C

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integral of (x^2+3x-1)^4(2x+3)

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

\frac{1}{5120} (4 x^{2} + 12 x - 4)^{5} + C

Solution steps

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

Apply u-substitution

= \int \frac{u ( u ^{2} - 13 ) ^{4}}{512} d u

Take the constant out:

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

= \frac{1}{512} \cdot \int u (u^{2} - 13)^{4} d u

Apply u-substitution

= \frac{1}{512} \cdot \int \frac{v ^{4}}{2} d v

Take the constant out:

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

= \frac{1}{512} \cdot \frac{1}{2} \cdot \int v^{4} d v

Apply the Power Rule

= \frac{1}{512} \cdot \frac{1}{2} \cdot \frac{v ^{5}}{5}

Substitute back

= \frac{1}{512} \cdot \frac{1}{2} \cdot \frac{( ( 2 x + 3 ) ^{2} - 13 ) ^{5}}{5}

Simplify

\frac{1}{512} \cdot \frac{1}{2} \cdot \frac{( ( 2 x + 3 ) ^{2} - 13 ) ^{5}}{5} : \frac{1}{5120} (4 x^{2} + 12 x - 4)^{5}

= \frac{1}{5120} (4 x^{2} + 12 x - 4)^{5}

Add a constant to the solution

= \frac{1}{5120} (4 x^{2} + 12 x - 4)^{5} + C

What is the integral of (x^2+3x-1)^4(2x+3) ?
The integral of (x^2+3x-1)^4(2x+3) is 1/5120 (4x^2+12x-4)^5+C