What is the integral of (x^3)/(sqrt((2x^2+1)^3)) ?

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

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integral of (x^3)/(sqrt((2x^2+1)^3))

\int \frac{x ^{3}}{( 2 x ^{2} + 1 ) ^{3}} d x

\frac{x ^{2} + 1}{2 2 x ^{2} + 1} + C

Solution steps

\int \frac{x ^{3}}{( 2 x ^{2} + 1 ) ^{3}} d x

Apply u-substitution

: \int \frac{u - 1}{8 u ^{\frac{3}{2}}} d u

= \int \frac{u - 1}{8 u ^{\frac{3}{2}}} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{u - 1}{u ^{\frac{3}{2}}} d u

Expand

\frac{u - 1}{u ^{\frac{3}{2}}} : \frac{1}{u ^{\frac{1}{2}}} - \frac{1}{u ^{\frac{3}{2}}}

= \frac{1}{8} \cdot \int \frac{1}{u ^{\frac{1}{2}}} - \frac{1}{u ^{\frac{3}{2}}} 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}{8} (\int \frac{1}{u ^{\frac{1}{2}}} d u - \int \frac{1}{u ^{\frac{3}{2}}} d u)

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

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

= \frac{1}{8} (2 u^{\frac{1}{2}} - (- \frac{2}{u}))

Substitute back

u = 2 x^{2} + 1

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

Simplify

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

= \frac{x ^{2} + 1}{2 2 x ^{2} + 1}

Add a constant to the solution

= \frac{x ^{2} + 1}{2 2 x ^{2} + 1} + C

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