What is the integral of (3y^4+3y^2-1)/(y^3+y) ?

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

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integral of (3y^4+3y^2-1)/(y^3+y)

\int \frac{3 y ^{4} + 3 y ^{2} - 1}{y ^{3} + y} d y

\frac{3 y ^{2}}{2} - ln ∣ y ∣ + \frac{1}{2} ln y^{2} + 1 + C

Solution steps

\int \frac{3 y ^{4} + 3 y ^{2} - 1}{y ^{3} + y} d y

Take the partial fraction of

\frac{3 y ^{4} + 3 y ^{2} - 1}{y ^{3} + y} : 3 y - \frac{1}{y} + \frac{y}{y ^{2} + 1}

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

Apply the Sum Rule:

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

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

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

\int \frac{1}{y} d y = ln ∣ y ∣

\int \frac{y}{y ^{2} + 1} d y = \frac{1}{2} ln y^{2} + 1

= \frac{3 y ^{2}}{2} - ln ∣ y ∣ + \frac{1}{2} ln y^{2} + 1

Add a constant to the solution

= \frac{3 y ^{2}}{2} - ln ∣ y ∣ + \frac{1}{2} ln y^{2} + 1 + C

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x \to 10 lim (15)

36 y^{^{''}} + 12 y^{^{'}} + y = 0

\int_{- 2 π}^{2 π} x^{2} sin (x) d x

\int sec (x) cos (x) d x

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