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

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

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integral of x^2sqrt(x+5)

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

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

Solution steps

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

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

Add a constant to the solution

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

\int in y d y

(1 + x^{4}) \frac{d y}{d x} = \frac{x ^{3}}{y}

\int 3 e^{\frac{3 t}{2}} t d t

f (x) = 4 x + 6 x, at x = 4

f (x) = - 3 - 4 x^{2}, (5, - 103)

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