What is the integral of (x^3)/(sqrt(x^2+\sqrt{2))} ?

The integral of (x^3)/(sqrt(x^2+\sqrt{2))} is

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integral of (x^3)/(sqrt(x^2+\sqrt{2))}

\int \frac{x ^{3}}{x ^{2} + 2} d x

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

Solution steps

\int \frac{x ^{3}}{x ^{2} + 2} d x

Apply u-substitution

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

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

Expand

\frac{u - 2}{u} : u - \frac{2}{u}

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

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

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

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

Substitute back

u = x^{2} + 2

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{cos ( a x )}{1 - sin ( a x )} d x

n = 1 \sum \infty \frac{x ^{n}}{4 n - 1}

y = e^{(4 x + 3)}

\frac{∣ x ∣}{2 - x ^{2}}

(1 + x^{2}) y^{^{'}} + 2 x y = \frac{1}{( 1 + x ^{2} ) y}

What is the integral of (x^3)/(sqrt(x^2+\sqrt{2))} ?
The integral of (x^3)/(sqrt(x^2+\sqrt{2))} is