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

The integral of x(2+x^2)^3 is 1/8 (2+x^2)^4+C

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integral of x(2+x^2)^3

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

\frac{1}{8} (2 + x^{2})^{4} + C

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: \frac{u ^{4}}{4}

= \frac{1}{2} \cdot \frac{u ^{4}}{4}

Substitute back

u = 2 + x^{2}

= \frac{1}{2} \cdot \frac{( 2 + x ^{2} ) ^{4}}{4}

Simplify

\frac{1}{2} \cdot \frac{( 2 + x ^{2} ) ^{4}}{4} : \frac{1}{8} (2 + x^{2})^{4}

= \frac{1}{8} (2 + x^{2})^{4}

Add a constant to the solution

= \frac{1}{8} (2 + x^{2})^{4} + C

\frac{- a - b}{( s + 1 )}

\frac{2}{3} x^{\frac{3}{2}}

\frac{3}{s ( s + 2 )}

\int \frac{1}{8 x} d x

y^{^{''}} - 5 y^{^{'}} + 6 y = 36 t

What is the integral of x(2+x^2)^3 ?
The integral of x(2+x^2)^3 is 1/8 (2+x^2)^4+C