What is the integral of (x^3+ex)sqrt(x^2+e) ?

The integral of (x^3+ex)sqrt(x^2+e) is 1/5 (x^2+e)^{5/2}+C

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integral of (x^3+ex)sqrt(x^2+e)

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

\frac{1}{5} (x^{2} + e)^{\frac{5}{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{u u}{2} d u

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

Take the constant out:

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

= \frac{1}{2} \cdot \int u u d u

Simplify

u u : u^{\frac{3}{2}}

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

Apply the Power Rule

: \frac{2}{5} u^{\frac{5}{2}}

= \frac{1}{2} \cdot \frac{2}{5} u^{\frac{5}{2}}

Substitute back

u = x^{2} + e

= \frac{1}{2} \cdot \frac{2}{5} (x^{2} + e)^{\frac{5}{2}}

Simplify

\frac{1}{2} \cdot \frac{2}{5} (x^{2} + e)^{\frac{5}{2}} : \frac{1}{5} (x^{2} + e)^{\frac{5}{2}}

= \frac{1}{5} (x^{2} + e)^{\frac{5}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} + e)^{\frac{5}{2}} + C

- 4 x sin (2 x^{2})

f (x) = - 8 (4 x + 3)^{- 3}, at x = 3

\frac{d}{d x} (cot (x) + csc (x))

x \to 1 lim (\frac{1 - x ^{2}}{ln ( x )})

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