What is the integral of ((y^3+4y))/(9+y^4) ?

The integral of ((y^3+4y))/(9+y^4) is 1/4 ln|9+y^4|+2/3 arctan((y^2)/3)+C

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integral of ((y^3+4y))/(9+y^4)

\int \frac{( y ^{3} + 4 y )}{9 + y ^{4}} d y

\frac{1}{4} ln 9 + y^{4} + \frac{2}{3} arctan (\frac{y ^{2}}{3}) + C

Solution steps

\int \frac{y ^{3} + 4 y}{9 + y ^{4}} d y

Expand

\frac{y ^{3} + 4 y}{9 + y ^{4}} : \frac{y ^{3}}{9 + y ^{4}} + \frac{4 y}{9 + y ^{4}}

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{y ^{3}}{9 + y ^{4}} d y + \int \frac{4 y}{9 + y ^{4}} d y

\int \frac{y ^{3}}{9 + y ^{4}} d y = \frac{1}{4} ln 9 + y^{4}

\int \frac{4 y}{9 + y ^{4}} d y = \frac{2}{3} arctan (\frac{y ^{2}}{3})

= \frac{1}{4} ln 9 + y^{4} + \frac{2}{3} arctan (\frac{y ^{2}}{3})

Add a constant to the solution

= \frac{1}{4} ln 9 + y^{4} + \frac{2}{3} arctan (\frac{y ^{2}}{3}) + C

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What is the integral of ((y^3+4y))/(9+y^4) ?
The integral of ((y^3+4y))/(9+y^4) is 1/4 ln|9+y^4|+2/3 arctan((y^2)/3)+C