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

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

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integral of xsqrt(y+x)

\int x y + x d x

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

Solution steps

\int x y + x d x

Apply u-substitution

: \int (u - y) u d u

= \int (u - y) u d u

Expand

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

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

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

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

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

Substitute back

u = y + x

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

Add a constant to the solution

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

\int x^{2} ln (2 x) d x

\frac{d}{d x} (arccos (ln (x)))

\frac{d}{d t} (3 sin (2 t))

\int_{3}^{5} x^{2} d x

\int \frac{t ^{3}}{t ^{3} + 1} d t

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