What is the integral of 1/(x^3+xy^2) ?

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

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integral of 1/(x^3+xy^2)

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

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

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{x ^{2} ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{x ^{2}} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{x ^{2}} arctan (u)

Substitute back

u = \frac{x}{x x} y

= \frac{1}{x ^{2}} arctan (\frac{x}{x x} y)

Simplify

\frac{1}{x ^{2}} arctan (\frac{x}{x x} y) : \frac{1}{x ^{2}} arctan (\frac{y}{x})

= \frac{1}{x ^{2}} arctan (\frac{y}{x})

Add a constant to the solution

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

\frac{d}{d x} (- 16 sin (4 x))

\frac{d}{d x} (ln (\frac{1 + x}{x ^{3}}))

\int_{\frac{1}{5}}^{7} 8 x ln (5 x) d x

y^{^{'}} + \frac{2}{x} y = 4 x

\frac{d}{d x} (sin^{2} (\frac{x}{2}))

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