What is the integral of (10)/(36*x^4-13*x^2+1) ?

The integral of (10)/(36*x^4-13*x^2+1) is 3ln|3x+1|-3ln|3x-1|-2ln|2x+1|+2ln|2x-1|+C

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integral of (10)/(36x^4-13x^2+1)

\int \frac{10}{36 \cdot x ^{4} - 13 \cdot x ^{2} + 1} d x

3 ln ∣ 3 x + 1 ∣ - 3 ln ∣ 3 x - 1 ∣ - 2 ln ∣ 2 x + 1 ∣ + 2 ln ∣ 2 x - 1 ∣ + C

Solution steps

\int \frac{10}{36 x ^{4} - 13 x ^{2} + 1} d x

Take the constant out:

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

= 10 \cdot \int \frac{1}{36 x ^{4} - 13 x ^{2} + 1} d x

Take the partial fraction of

\frac{1}{36 x ^{4} - 13 x ^{2} + 1} : \frac{9}{10 ( 3 x + 1 )} - \frac{9}{10 ( 3 x - 1 )} - \frac{2}{5 ( 2 x + 1 )} + \frac{2}{5 ( 2 x - 1 )}

= 10 \cdot \int \frac{9}{10 ( 3 x + 1 )} - \frac{9}{10 ( 3 x - 1 )} - \frac{2}{5 ( 2 x + 1 )} + \frac{2}{5 ( 2 x - 1 )} d x

Apply the Sum Rule:

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

= 10 (\int \frac{9}{10 ( 3 x + 1 )} d x - \int \frac{9}{10 ( 3 x - 1 )} d x - \int \frac{2}{5 ( 2 x + 1 )} d x + \int \frac{2}{5 ( 2 x - 1 )} d x)

\int \frac{9}{10 ( 3 x + 1 )} d x = \frac{3}{10} ln ∣ 3 x + 1 ∣

\int \frac{9}{10 ( 3 x - 1 )} d x = \frac{3}{10} ln ∣ 3 x - 1 ∣

\int \frac{2}{5 ( 2 x + 1 )} d x = \frac{1}{5} ln ∣ 2 x + 1 ∣

\int \frac{2}{5 ( 2 x - 1 )} d x = \frac{1}{5} ln ∣ 2 x - 1 ∣

= 10 (\frac{3}{10} ln ∣ 3 x + 1 ∣ - \frac{3}{10} ln ∣ 3 x - 1 ∣ - \frac{1}{5} ln ∣ 2 x + 1 ∣ + \frac{1}{5} ln ∣ 2 x - 1 ∣)

Simplify

10 (\frac{3}{10} ln ∣ 3 x + 1 ∣ - \frac{3}{10} ln ∣ 3 x - 1 ∣ - \frac{1}{5} ln ∣ 2 x + 1 ∣ + \frac{1}{5} ln ∣ 2 x - 1 ∣) : 3 ln ∣ 3 x + 1 ∣ - 3 ln ∣ 3 x - 1 ∣ - 2 ln ∣ 2 x + 1 ∣ + 2 ln ∣ 2 x - 1 ∣

= 3 ln ∣ 3 x + 1 ∣ - 3 ln ∣ 3 x - 1 ∣ - 2 ln ∣ 2 x + 1 ∣ + 2 ln ∣ 2 x - 1 ∣

Add a constant to the solution

= 3 ln ∣ 3 x + 1 ∣ - 3 ln ∣ 3 x - 1 ∣ - 2 ln ∣ 2 x + 1 ∣ + 2 ln ∣ 2 x - 1 ∣ + C

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What is the integral of (10)/(36*x^4-13*x^2+1) ?
The integral of (10)/(36*x^4-13*x^2+1) is 3ln|3x+1|-3ln|3x-1|-2ln|2x+1|+2ln|2x-1|+C