What is the integral of (x+1)sqrt(x+3) ?

The integral of (x+1)sqrt(x+3) is 2/5 (x+3)^{5/2}-4/3 (x+3)^{3/2}+C

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integral of (x+1)sqrt(x+3)

\int (x + 1) x + 3 d x

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

Solution steps

\int (x + 1) x + 3 d x

Apply u-substitution

: \int u (u - 2) d u

= \int u (u - 2) d u

Expand

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

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

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

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

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

Substitute back

u = x + 3

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

Add a constant to the solution

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

\frac{d y}{d x} = 4 x^{3} y

\frac{\partial}{\partial x} (- 2 y sin (x))

y^{^{''}} + 4 y = e^{- t}

7 (4 e^{x} - 27)^{5}

\int_{1}^{5} 1 + 2^{2} d x

What is the integral of (x+1)sqrt(x+3) ?
The integral of (x+1)sqrt(x+3) is 2/5 (x+3)^{5/2}-4/3 (x+3)^{3/2}+C