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

The integral of x^2sqrt(3+x) is 2/7 (3+x)^{7/2}-12/5 (3+x)^{5/2}+6(3+x)^{3/2}+C

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integral of x^2sqrt(3+x)

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

\frac{2}{7} (3 + x)^{\frac{7}{2}} - \frac{12}{5} (3 + x)^{\frac{5}{2}} + 6 (3 + x)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int (u - 3)^{2} u d u

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

Expand

(u - 3)^{2} u : u^{\frac{5}{2}} - 6 u^{\frac{3}{2}} + 9 u

= \int u^{\frac{5}{2}} - 6 u^{\frac{3}{2}} + 9 u 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

= \int u^{\frac{5}{2}} d u - \int 6 u^{\frac{3}{2}} d u + \int 9 u d u

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 6 u^{\frac{3}{2}} d u = \frac{12}{5} u^{\frac{5}{2}}

\int 9 u d u = 6 u^{\frac{3}{2}}

= \frac{2}{7} u^{\frac{7}{2}} - \frac{12}{5} u^{\frac{5}{2}} + 6 u^{\frac{3}{2}}

Substitute back

u = 3 + x

= \frac{2}{7} (3 + x)^{\frac{7}{2}} - \frac{12}{5} (3 + x)^{\frac{5}{2}} + 6 (3 + x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{7} (3 + x)^{\frac{7}{2}} - \frac{12}{5} (3 + x)^{\frac{5}{2}} + 6 (3 + x)^{\frac{3}{2}} + C

y = \frac{e ^{x} + 1}{e ^{x} - 1}

\frac{\partial}{\partial x} (- \frac{y}{x ^{2}} + 4 x + \frac{1}{y})

x^{2} \frac{d y}{d x} - 2 x y = 5 y^{4}

\int_{- 2}^{2} (x + 2) 4 - x^{2} d x

y^{^{''}} + 12 y^{^{'}} + 40 y = 0

What is the integral of x^2sqrt(3+x) ?
The integral of x^2sqrt(3+x) is 2/7 (3+x)^{7/2}-12/5 (3+x)^{5/2}+6(3+x)^{3/2}+C