What is the integral of (x^3+1)/5 (x^3-3x^2+5) ?

The integral of (x^3+1)/5 (x^3-3x^2+5) is 1/70 (2x^7-7x^6+21x^4-14x^3+70x)+C

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integral of (x^3+1)/5 (x^3-3x^2+5)

\int \frac{x ^{3} + 1}{5} (x^{3} - 3 x^{2} + 5) d x

\frac{1}{70} (2 x^{7} - 7 x^{6} + 21 x^{4} - 14 x^{3} + 70 x) + C

Solution steps

\int \frac{x ^{3} + 1}{5} (x^{3} - 3 x^{2} + 5) d x

Take the constant out:

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

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

Expand

(x^{3} + 1) (x^{3} - 3 x^{2} + 5) : x^{6} - 3 x^{5} + 6 x^{3} - 3 x^{2} + 5

= \frac{1}{5} \cdot \int x^{6} - 3 x^{5} + 6 x^{3} - 3 x^{2} + 5 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

= \frac{1}{5} (\int x^{6} d x - \int 3 x^{5} d x + \int 6 x^{3} d x - \int 3 x^{2} d x + \int 5 d x)

\int x^{6} d x = \frac{x ^{7}}{7}

\int 3 x^{5} d x = \frac{x ^{6}}{2}

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

\int 3 x^{2} d x = x^{3}

\int 5 d x = 5 x

= \frac{1}{5} (\frac{x ^{7}}{7} - \frac{x ^{6}}{2} + \frac{3 x ^{4}}{2} - x^{3} + 5 x)

Simplify

\frac{1}{5} (\frac{x ^{7}}{7} - \frac{x ^{6}}{2} + \frac{3 x ^{4}}{2} - x^{3} + 5 x) : \frac{1}{70} (2 x^{7} - 7 x^{6} + 21 x^{4} - 14 x^{3} + 70 x)

= \frac{1}{70} (2 x^{7} - 7 x^{6} + 21 x^{4} - 14 x^{3} + 70 x)

Add a constant to the solution

= \frac{1}{70} (2 x^{7} - 7 x^{6} + 21 x^{4} - 14 x^{3} + 70 x) + C

\int h^{3} e^{h^{2}} d h

\int (3 x + 7) d x

\int \frac{x ^{4} - 6}{x ^{3} - 2 x ^{2}} d x

\int (x^{a} - x^{- a})^{2} d x

What is the integral of (x^3+1)/5 (x^3-3x^2+5) ?
The integral of (x^3+1)/5 (x^3-3x^2+5) is 1/70 (2x^7-7x^6+21x^4-14x^3+70x)+C