What is the integral of ysqrt(y+1) ?

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

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integral of ysqrt(y+1)

\int y y + 1 d y

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

Solution steps

\int y y + 1 d y

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 = y + 1

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

Add a constant to the solution

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

d er i v a t i v e 4 tan^{2} (3 x)

\int - ln (x) d x

\frac{\partial}{\partial y} (\frac{1}{x ^{2} + y ^{2}})

n \to \infty lim (an)

x \to 3 lim (\frac{7}{x - 3})

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