What is the integral of ysqrt(x+y^2) ?

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

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integral of ysqrt(x+y^2)

\int y x + y^{2} d x

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

Solution steps

\int y x + y^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= y \cdot \int y^{2} + x d x

Apply Integration By Parts

: x y^{2} + x - \int \frac{x}{2 y ^{2} + x} d x

= y (x y^{2} + x - \int \frac{x}{2 y ^{2} + x} d x)

\int \frac{x}{2 y ^{2} + x} d x = \frac{1}{3} (y^{2} + x)^{\frac{3}{2}} - y^{2} y^{2} + x

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

Simplify

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

Add a constant to the solution

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

\int (3 sec (x) - cos (x))^{2} d x

x^{2} - 4 x + 39

\int_{1}^{4} π (\frac{1}{x})^{2} d x

y^{^{'}} = 3 x^{2} y

x \to 0 + lim (\frac{ln ( x )}{csc ( x )})

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