What is the integral of 2x^7sqrt(x^4+1) ?

The integral of 2x^7sqrt(x^4+1) is 1/15 (3x^4-2)(x^4+1)^{3/2}+C

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integral of 2x^7sqrt(x^4+1)

\int 2 x^{7} x^{4} + 1 d x

\frac{1}{15} (3 x^{4} - 2) (x^{4} + 1)^{\frac{3}{2}} + C

Solution steps

\int 2 x^{7} x^{4} + 1 d x

Take the constant out:

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

= 2 \cdot \int x^{7} x^{4} + 1 d x

Apply u-substitution

: \int \frac{u ^{\frac{1}{7}} u ^{\frac{4}{7}} + 1}{7} d u

= 2 \cdot \int \frac{u ^{\frac{1}{7}} u ^{\frac{4}{7}} + 1}{7} d u

Take the constant out:

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

= 2 \cdot \frac{1}{7} \cdot \int u^{\frac{1}{7}} u^{\frac{4}{7}} + 1 d u

Apply u-substitution

: \int \frac{7 ( v - 1 ) v}{4} d v

= 2 \cdot \frac{1}{7} \cdot \int \frac{7 ( v - 1 ) v}{4} d v

Take the constant out:

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

= 2 \cdot \frac{1}{7} \cdot \frac{7}{4} \cdot \int (v - 1) v d v

Expand

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

= 2 \cdot \frac{1}{7} \cdot \frac{7}{4} \cdot \int v^{\frac{3}{2}} - v d v

Apply the Sum Rule:

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

= 2 \cdot \frac{1}{7} \cdot \frac{7}{4} (\int v^{\frac{3}{2}} d v - \int v d v)

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

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

= 2 \cdot \frac{1}{7} \cdot \frac{7}{4} (\frac{2}{5} v^{\frac{5}{2}} - \frac{2}{3} v^{\frac{3}{2}})

Substitute back

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

Simplify

2 \cdot \frac{1}{7} \cdot \frac{7}{4} (\frac{2}{5} ((x^{7})^{\frac{4}{7}} + 1)^{\frac{5}{2}} - \frac{2}{3} ((x^{7})^{\frac{4}{7}} + 1)^{\frac{3}{2}}) : \frac{1}{15} (3 x^{4} - 2) (x^{4} + 1)^{\frac{3}{2}}

= \frac{1}{15} (3 x^{4} - 2) (x^{4} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{15} (3 x^{4} - 2) (x^{4} + 1)^{\frac{3}{2}} + C

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What is the integral of 2x^7sqrt(x^4+1) ?
The integral of 2x^7sqrt(x^4+1) is 1/15 (3x^4-2)(x^4+1)^{3/2}+C