What is the integral of (2xy^5)/(sqrt(9+x^2y^2-y^4)) ?

The integral of (2xy^5)/(sqrt(9+x^2y^2-y^4)) is 2y^3sqrt(y^2x^2+9-y^4)+C

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integral of (2xy^5)/(sqrt(9+x^2y^2-y^4))

\int \frac{2 x y ^{5}}{9 + x ^{2} y ^{2} - y ^{4}} d x

2 y^{3} y^{2} x^{2} + 9 - y^{4} + C

Solution steps

\int \frac{2 x y ^{5}}{9 + x ^{2} y ^{2} - y ^{4}} d x

Take the constant out:

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

= 2 y^{5} \cdot \int \frac{x}{9 - y ^{4} + y ^{2} x ^{2}} d x

Apply u-substitution

: \int \frac{1}{2 y ^{2} u} d u

= 2 y^{5} \cdot \int \frac{1}{2 y ^{2} u} d u

Take the constant out:

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

= 2 y^{5} \frac{1}{2 y ^{2}} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= 2 y^{5} \frac{1}{2 y ^{2}} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 9 - y^{4} + y^{2} x^{2}

= 2 y^{5} \frac{1}{2 y ^{2}} \cdot 2 (9 - y^{4} + y^{2} x^{2})^{\frac{1}{2}}

Simplify

2 y^{5} \frac{1}{2 y ^{2}} \cdot 2 (9 - y^{4} + y^{2} x^{2})^{\frac{1}{2}} : 2 y^{3} y^{2} x^{2} + 9 - y^{4}

= 2 y^{3} y^{2} x^{2} + 9 - y^{4}

Add a constant to the solution

= 2 y^{3} y^{2} x^{2} + 9 - y^{4} + C

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What is the integral of (2xy^5)/(sqrt(9+x^2y^2-y^4)) ?
The integral of (2xy^5)/(sqrt(9+x^2y^2-y^4)) is 2y^3sqrt(y^2x^2+9-y^4)+C