What is the integral of sqrt(x)sqrt(x\sqrt{x)+1} ?

The integral of sqrt(x)sqrt(x\sqrt{x)+1} is 4/9 (xsqrt(x)+1)^{3/2}+C

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integral of sqrt(x)sqrt(x\sqrt{x)+1}

\int x x x + 1 d x

\frac{4}{9} (x x + 1)^{\frac{3}{2}} + C

Solution steps

\int x x x + 1 d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

= \frac{2}{3} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = x x + 1

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

Simplify

\frac{2}{3} \cdot \frac{2}{3} (x x + 1)^{\frac{3}{2}} : \frac{4}{9} (x x + 1)^{\frac{3}{2}}

= \frac{4}{9} (x x + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{4}{9} (x x + 1)^{\frac{3}{2}} + C

y^{^{''}} + 4 y^{^{'}} + 3 y = 15 t e^{3 t}

x^{2} = 4 y, 3 x - 4 y + 4 = 0

y^{^{'}} + y = e^{- t} sin (t), y (0) = 1

y^{^{'}} = y (1 - y)

x \to - 4 lim (8 x^{2} + 1)

What is the integral of sqrt(x)sqrt(x\sqrt{x)+1} ?
The integral of sqrt(x)sqrt(x\sqrt{x)+1} is 4/9 (xsqrt(x)+1)^{3/2}+C