What is the integral of (x^2y^3-1/(1+9x^2)) ?

The integral of (x^2y^3-1/(1+9x^2)) is -1/3 arctan(3x)+(y^3x^3)/3+C

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integral of (x^2y^3-1/(1+9x^2))

\int (x^{2} y^{3} - \frac{1}{1 + 9 x ^{2}}) d x

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

Solution steps

\int x^{2} y^{3} - \frac{1}{1 + 9 x ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{1}{1 + 9 x ^{2}} d x + \int y^{3} x^{2} d x

\int \frac{1}{1 + 9 x ^{2}} d x = \frac{1}{3} arctan (3 x)

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

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

Add a constant to the solution

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

\int \frac{14 x - 13}{x ^{2} - 4 x - 21} d x

\frac{\partial}{\partial x} (\frac{7 y}{x ^{2} + y ^{2} + 19})

- \frac{18}{( x - 9 ) ^{2}}

f (x) = 22 x

y = - 6 (4 x^{2} + 4)^{- 6}

What is the integral of (x^2y^3-1/(1+9x^2)) ?
The integral of (x^2y^3-1/(1+9x^2)) is -1/3 arctan(3x)+(y^3x^3)/3+C