What is the integral of x^7sqrt(x^4+3) ?

The integral of x^7sqrt(x^4+3) is 1/10 (x^4-2)(x^4+3)^{3/2}+C

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integral of x^7sqrt(x^4+3)

\int x^{7} x^{4} + 3 d x

\frac{1}{10} (x^{4} - 2) (x^{4} + 3)^{\frac{3}{2}} + C

Solution steps

\int x^{7} x^{4} + 3 d x

Apply u-substitution

: \int \frac{( u - 3 ) u}{4} d u

= \int \frac{( u - 3 ) u}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int (u - 3) u d u

Expand

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

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

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

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

= \frac{1}{4} (\frac{2}{5} u^{\frac{5}{2}} - 2 u^{\frac{3}{2}})

Substitute back

u = x^{4} + 3

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

Simplify

\frac{1}{4} (\frac{2}{5} (x^{4} + 3)^{\frac{5}{2}} - 2 (x^{4} + 3)^{\frac{3}{2}}) : \frac{1}{10} (x^{4} - 2) (x^{4} + 3)^{\frac{3}{2}}

= \frac{1}{10} (x^{4} - 2) (x^{4} + 3)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{10} (x^{4} - 2) (x^{4} + 3)^{\frac{3}{2}} + C

f (x) = \frac{3}{x + 1}, (2, 1)

y^{^{'}} = e^{- 3 x} + 2 x

\frac{d}{d x} (x (x^{2} + 1)^{10})

\int - \frac{1}{Ux} + \frac{1}{U ( x - U )} d x

\frac{\partial}{\partial y} (\frac{e ^{y}}{x + y ^{2}})

What is the integral of x^7sqrt(x^4+3) ?
The integral of x^7sqrt(x^4+3) is 1/10 (x^4-2)(x^4+3)^{3/2}+C