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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int u^{2} - 1 d u

= 2 \cdot \int u^{2} - 1 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

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

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

\int 1 d u = u

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

Substitute back

u = 1 + x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{2 x ^{2} + 4 x}{x ^{2} + 4 x + 8} d x

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

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