What is the integral of xsqrt(4+x) ?

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

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integral of xsqrt(4+x)

\int x 4 + x d x

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

Solution steps

\int x 4 + x d x

Apply u-substitution

: \int (u - 4) u d u

= \int (u - 4) u d u

Expand

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

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

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

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

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

Substitute back

u = 4 + x

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

Add a constant to the solution

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

f (x) = x^{3} - 3 x^{2} + 5, at x = - 1

\int_{0}^{6} 2

\int_{1}^{100} \frac{13}{x ^{3}} d x

f (t) = t e^{t}

x \to 0 lim (x^{4} \cdot cos (\frac{π}{x}))

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