What is the integral of (2x^5+3x^2)/(sqrt(1+2x^3)) ?

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

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integral of (2x^5+3x^2)/(sqrt(1+2x^3))

\int \frac{2 x ^{5} + 3 x ^{2}}{1 + 2 x ^{3}} d x

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

Solution steps

\int \frac{2 x ^{5} + 3 x ^{2}}{1 + 2 x ^{3}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

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

\int 2 d u = 2 u

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

Substitute back

u = 1 + 2 x^{3}

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

Simplify

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

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

Add a constant to the solution

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

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