What is the integral of sqrt(x(1+\sqrt{x))}^4 ?

The integral of sqrt(x(1+\sqrt{x))}^4 is (x^3)/3+4/7 x^{7/2}+(x^4)/4+C

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integral of sqrt(x(1+\sqrt{x))}^4

\int x (1 + x)^{4} d x

\frac{x ^{3}}{3} + \frac{4}{7} x^{\frac{7}{2}} + \frac{x ^{4}}{4} + C

Solution steps

\int (x (1 + x))^{4} d x

Expand

(x (1 + x))^{4} : x^{2} + 2 x^{\frac{5}{2}} + x^{3}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 2 x^{\frac{5}{2}} d x = \frac{4}{7} x^{\frac{7}{2}}

\int x^{3} d x = \frac{x ^{4}}{4}

= \frac{x ^{3}}{3} + \frac{4}{7} x^{\frac{7}{2}} + \frac{x ^{4}}{4}

Add a constant to the solution

= \frac{x ^{3}}{3} + \frac{4}{7} x^{\frac{7}{2}} + \frac{x ^{4}}{4} + C

y^{^{'}} = t - y t

\frac{d y}{d x}, y = 2 x + 1

\int sin (2 \cdot x) d x

\int 15 - 14 x - x^{2} d x

(y + x^{3} + x y^{2}) d x - (x) d y = 0

What is the integral of sqrt(x(1+\sqrt{x))}^4 ?
The integral of sqrt(x(1+\sqrt{x))}^4 is (x^3)/3+4/7 x^{7/2}+(x^4)/4+C