What is the integral of x^{13}sqrt(x^7+1) ?

The integral of x^{13}sqrt(x^7+1) is 1/7 (2/5 (x^7+1)^{5/2}-2/3 (x^7+1)^{3/2})+C

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integral of x^{13}sqrt(x^7+1)

\int x^{13} x^{7} + 1 d x

\frac{1}{7} (\frac{2}{5} (x^{7} + 1)^{\frac{5}{2}} - \frac{2}{3} (x^{7} + 1)^{\frac{3}{2}}) + C

Solution steps

\int x^{13} x^{7} + 1 d x

Apply u-substitution

= \int \frac{( u - 1 ) u}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int (u - 1) u d u

Expand

(u - 1) u : u^{\frac{3}{2}} - u

= \frac{1}{7} \cdot \int u^{\frac{3}{2}} - u d u

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}{7} (\int u^{\frac{3}{2}} d u - \int u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int u d u = \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{7} (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = x^{7} + 1

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

Add a constant to the solution

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

What is the integral of x^{13}sqrt(x^7+1) ?
The integral of x^{13}sqrt(x^7+1) is 1/7 (2/5 (x^7+1)^{5/2}-2/3 (x^7+1)^{3/2})+C