What is the integral of sqrt(3+x)(x+1)^2 ?

The integral of sqrt(3+x)(x+1)^2 is 2/7 (x+3)^{7/2}-8/5 (x+3)^{5/2}+8/3 (x+3)^{3/2}+C

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integral of sqrt(3+x)(x+1)^2

\int 3 + x (x + 1)^{2} d x

\frac{2}{7} (x + 3)^{\frac{7}{2}} - \frac{8}{5} (x + 3)^{\frac{5}{2}} + \frac{8}{3} (x + 3)^{\frac{3}{2}} + C

Solution steps

\int 3 + x (x + 1)^{2} d x

Apply u-substitution

: \int u^{2} u + 2 d u

= \int u^{2} u + 2 d u

Apply u-substitution

: \int (v - 2)^{2} v d v

= \int (v - 2)^{2} v d v

Expand

(v - 2)^{2} v : v^{\frac{5}{2}} - 4 v^{\frac{3}{2}} + 4 v

= \int v^{\frac{5}{2}} - 4 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{5}{2}} d v - \int 4 v^{\frac{3}{2}} d v + \int 4 v d v

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

\int 4 v^{\frac{3}{2}} d v = \frac{8}{5} v^{\frac{5}{2}}

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

= \frac{2}{7} v^{\frac{7}{2}} - \frac{8}{5} v^{\frac{5}{2}} + \frac{8}{3} v^{\frac{3}{2}}

Substitute back

= \frac{2}{7} (x + 1 + 2)^{\frac{7}{2}} - \frac{8}{5} (x + 1 + 2)^{\frac{5}{2}} + \frac{8}{3} (x + 1 + 2)^{\frac{3}{2}}

Simplify

\frac{2}{7} (x + 1 + 2)^{\frac{7}{2}} - \frac{8}{5} (x + 1 + 2)^{\frac{5}{2}} + \frac{8}{3} (x + 1 + 2)^{\frac{3}{2}} : \frac{2}{7} (x + 3)^{\frac{7}{2}} - \frac{8}{5} (x + 3)^{\frac{5}{2}} + \frac{8}{3} (x + 3)^{\frac{3}{2}}

= \frac{2}{7} (x + 3)^{\frac{7}{2}} - \frac{8}{5} (x + 3)^{\frac{5}{2}} + \frac{8}{3} (x + 3)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{7} (x + 3)^{\frac{7}{2}} - \frac{8}{5} (x + 3)^{\frac{5}{2}} + \frac{8}{3} (x + 3)^{\frac{3}{2}} + C

y^{^{''}} + 2 y^{^{'}} - 3 y = t (e^{- 3 t} + e^{2 t})

e^{\frac{- y ^{2}}{2}}

\frac{d}{d x} (\frac{x - 2}{2})

\frac{d}{d x} (x^{5} tan (x))

n = 1 \sum \infty \frac{1}{n 3 ^{n}} x^{n}

What is the integral of sqrt(3+x)(x+1)^2 ?
The integral of sqrt(3+x)(x+1)^2 is 2/7 (x+3)^{7/2}-8/5 (x+3)^{5/2}+8/3 (x+3)^{3/2}+C