What is the integral of (15)/(3y^2-4y+4) ?

The integral of (15)/(3y^2-4y+4) is (15)/(4+3y^2-4y)x+C

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integral of (15)/(3y^2-4y+4)

\int \frac{15}{3 y ^{2} - 4 y + 4} d x

\frac{15}{4 + 3 y ^{2} - 4 y} x + C

Solution steps

\int \frac{15}{3 y ^{2} - 4 y + 4} d x

Integral of a constant:

\int a d x = a x

= \frac{15}{4 + 3 y ^{2} - 4 y} x

Add a constant to the solution

= \frac{15}{4 + 3 y ^{2} - 4 y} x + C

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What is the integral of (15)/(3y^2-4y+4) ?
The integral of (15)/(3y^2-4y+4) is (15)/(4+3y^2-4y)x+C