What is the integral of (3n-4)/(n^2-2n) ?

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

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integral of (3n-4)/(n^2-2n)

\int \frac{3 n - 4}{n ^{2} - 2 n}

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

Solution steps

\int \frac{3 n - 4}{n ^{2} - 2 n} d x

Integral of a constant:

\int a d x = a x

= \frac{3 n - 4}{n ^{2} - 2 n} x

Add a constant to the solution

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

\int ((1 + x y) e^{x y}) d x

y - x^{2} y = 2 - 2 x^{3}

\int \frac{7}{x ^{2} x ^{2} + 25} d x

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

\frac{d y}{d t} = 6 y^{2}

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