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

The integral of x^2(x^3+1)^{1/2} is 2/9 (x^3+1)^{3/2}+C

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integral of x^2(x^3+1)^{1/2}

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

\frac{2}{9} (x^{3} + 1)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

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

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

Multiply by the conjugate of

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

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

Apply u-substitution

: \int (v + 2)^{\frac{1}{2}} d v

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

Apply u-substitution

: \int w^{\frac{1}{2}} d w

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

Apply the Power Rule

: \frac{2}{3} w^{\frac{3}{2}}

= \frac{1}{3} \cdot \frac{2}{3} w^{\frac{3}{2}}

Substitute back

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

Simplify

\frac{1}{3} \cdot \frac{2}{3} (x^{3} - 1 + 2)^{\frac{3}{2}} : \frac{2}{9} (x^{3} + 1)^{\frac{3}{2}}

= \frac{2}{9} (x^{3} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{9} (x^{3} + 1)^{\frac{3}{2}} + C

c^{\frac{1}{2}}

\frac{- 2}{x} - \frac{- 4}{y} = 1

y = 3 cos (3 x), y = 3 - 3 cos (3 x), 0 \leq x \leq \frac{π}{3}

\frac{d}{d x} (e^{x^{3} + 4 x})

\int \frac{x ^{2} - 361}{x} d x

What is the integral of x^2(x^3+1)^{1/2} ?
The integral of x^2(x^3+1)^{1/2} is 2/9 (x^3+1)^{3/2}+C