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

The integral of x^2sqrt(5+x) is 2/7 (5+x)^{7/2}-4(5+x)^{5/2}+50/3 (5+x)^{3/2}+C

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integral of x^2sqrt(5+x)

\int x^{2} 5 + x d x

\frac{2}{7} (5 + x)^{\frac{7}{2}} - 4 (5 + x)^{\frac{5}{2}} + \frac{50}{3} (5 + x)^{\frac{3}{2}} + C

Solution steps

\int x^{2} 5 + x d x

Apply u-substitution

: \int (u - 5)^{2} u d u

= \int (u - 5)^{2} u d u

Expand

(u - 5)^{2} u : u^{\frac{5}{2}} - 10 u^{\frac{3}{2}} + 25 u

= \int u^{\frac{5}{2}} - 10 u^{\frac{3}{2}} + 25 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 10 u^{\frac{3}{2}} d u + \int 25 u d u

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

\int 10 u^{\frac{3}{2}} d u = 4 u^{\frac{5}{2}}

\int 25 u d u = \frac{50}{3} u^{\frac{3}{2}}

= \frac{2}{7} u^{\frac{7}{2}} - 4 u^{\frac{5}{2}} + \frac{50}{3} u^{\frac{3}{2}}

Substitute back

u = 5 + x

= \frac{2}{7} (5 + x)^{\frac{7}{2}} - 4 (5 + x)^{\frac{5}{2}} + \frac{50}{3} (5 + x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{7} (5 + x)^{\frac{7}{2}} - 4 (5 + x)^{\frac{5}{2}} + \frac{50}{3} (5 + x)^{\frac{3}{2}} + C

\frac{d y}{d x} = - 9 y

x \to - 3 lim (x (- 3))

\int 5 e^{x^{2}} d x

\int \frac{1}{x ^{2}} + 20 x^{\frac{1}{4}} d x

x \to 2 lim (x - 2)

What is the integral of x^2sqrt(5+x) ?
The integral of x^2sqrt(5+x) is 2/7 (5+x)^{7/2}-4(5+x)^{5/2}+50/3 (5+x)^{3/2}+C