What is the integral of 3x(x^2+8)^2 ?

The integral of 3x(x^2+8)^2 is 3((x^6)/6+4x^4+32x^2)+C

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integral of 3x(x^2+8)^2

\int 3 x (x^{2} + 8)^{2} d x

3 (\frac{x ^{6}}{6} + 4 x^{4} + 32 x^{2}) + C

Solution steps

\int 3 x (x^{2} + 8)^{2} d x

Take the constant out:

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

= 3 \cdot \int x (x^{2} + 8)^{2} d x

Expand

x (x^{2} + 8)^{2} : x^{5} + 16 x^{3} + 64 x

= 3 \cdot \int x^{5} + 16 x^{3} + 64 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

= 3 (\int x^{5} d x + \int 16 x^{3} d x + \int 64 x d x)

\int x^{5} d x = \frac{x ^{6}}{6}

\int 16 x^{3} d x = 4 x^{4}

\int 64 x d x = 32 x^{2}

= 3 (\frac{x ^{6}}{6} + 4 x^{4} + 32 x^{2})

Add a constant to the solution

= 3 (\frac{x ^{6}}{6} + 4 x^{4} + 32 x^{2}) + C

\int x^{2} (x^{3} + 8)^{10} d x

\int \frac{1}{( x + 3 ) x + 1} d x

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

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

\int \frac{e ^{3 x}}{3 - e ^{3 x}} d x

What is the integral of 3x(x^2+8)^2 ?
The integral of 3x(x^2+8)^2 is 3((x^6)/6+4x^4+32x^2)+C