What is the integral of usqrt(u+4) ?

The integral of usqrt(u+4) is 2/5 (u+4)^{5/2}-8/3 (u+4)^{3/2}+C

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integral of usqrt(u+4)

\int u u + 4 d u

\frac{2}{5} (u + 4)^{\frac{5}{2}} - \frac{8}{3} (u + 4)^{\frac{3}{2}} + C

Solution steps

\int u u + 4 d u

Apply u-substitution

: \int (v - 4) v d v

= \int (v - 4) v d v

Expand

(v - 4) v : v^{\frac{3}{2}} - 4 v

= \int v^{\frac{3}{2}} - 4 v d v

Apply the Sum Rule:

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

= \int v^{\frac{3}{2}} d v - \int 4 v d v

\int v^{\frac{3}{2}} d v = \frac{2}{5} v^{\frac{5}{2}}

\int 4 v d v = \frac{8}{3} v^{\frac{3}{2}}

= \frac{2}{5} v^{\frac{5}{2}} - \frac{8}{3} v^{\frac{3}{2}}

Substitute back

v = u + 4

= \frac{2}{5} (u + 4)^{\frac{5}{2}} - \frac{8}{3} (u + 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{5} (u + 4)^{\frac{5}{2}} - \frac{8}{3} (u + 4)^{\frac{3}{2}} + C

\frac{1}{( 3 + t ) ^{2}}

\frac{d}{d x} ((x - 5) (5 x + 4))

\frac{\partial}{\partial y} (x ln (x^{2} + y^{2}))

\int_{- 1}^{4} x - 3 x^{2} d x

(x e^{x^{2}})^{^{'}}

What is the integral of usqrt(u+4) ?
The integral of usqrt(u+4) is 2/5 (u+4)^{5/2}-8/3 (u+4)^{3/2}+C