What is the integral of (xsqrt(x+1)) ?

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

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integral of (xsqrt(x+1))

\int (x x + 1) d x

\frac{2}{5} (x + 1)^{\frac{5}{2}} - \frac{2}{3} (x + 1)^{\frac{3}{2}} + C

Solution steps

\int x x + 1 d x

Apply u-substitution

= \int (u - 1) u d u

Expand

(u - 1) u : u^{\frac{3}{2}} - u

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

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

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

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

Substitute back

u = x + 1

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

Add a constant to the solution

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

a re a 0.9 x^{2} + 6, x, (- 4, 4)

\frac{d y}{d x} = e^{2 x} + 4 y

\frac{d}{d x} (3 x + 2 x^{2})

x \to 6 lim (\frac{x ^{2} - 6}{x + 4})

\int \frac{19}{x ( x ^{2} + 4 ) ^{2}} d x

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