What is the integral of (4x^2-5)^28x ?

The integral of (4x^2-5)^28x is 8((8x^6)/3-10x^4+(25x^2)/2)+C

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integral of (4x^2-5)^28x

\int (4 x^{2} - 5)^{2} 8 x d x

8 (\frac{8 x ^{6}}{3} - 10 x^{4} + \frac{25 x ^{2}}{2}) + C

Solution steps

\int (4 x^{2} - 5)^{2} \cdot 8 x d x

Take the constant out:

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

= 8 \cdot \int (4 x^{2} - 5)^{2} x d x

Expand

(4 x^{2} - 5)^{2} x : 16 x^{5} - 40 x^{3} + 25 x

= 8 \cdot \int 16 x^{5} - 40 x^{3} + 25 x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 8 (\int 16 x^{5} d x - \int 40 x^{3} d x + \int 25 x d x)

\int 16 x^{5} d x = \frac{8 x ^{6}}{3}

\int 40 x^{3} d x = 10 x^{4}

\int 25 x d x = \frac{25 x ^{2}}{2}

= 8 (\frac{8 x ^{6}}{3} - 10 x^{4} + \frac{25 x ^{2}}{2})

Add a constant to the solution

= 8 (\frac{8 x ^{6}}{3} - 10 x^{4} + \frac{25 x ^{2}}{2}) + C

\int_{0}^{y} e^{- x} d x

\int (x + 3) ln (x) d x

\frac{d}{d x} (x^{4} y^{3})

\int sin (π) t d t

x^{2} y \cdot \frac{d y}{d x} - 9 = 0

What is the integral of (4x^2-5)^28x ?
The integral of (4x^2-5)^28x is 8((8x^6)/3-10x^4+(25x^2)/2)+C