What is the integral of (32x)/((2x-1)(4x^2-16x+15)) ?

The integral of (32x)/((2x-1)(4x^2-16x+15)) is ln|2x-1|-6ln|2x-3|+5ln|2x-5|+C

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integral of (32x)/((2x-1)(4x^2-16x+15))

\int \frac{32 x}{( 2 x - 1 ) ( 4 x ^{2} - 16 x + 15 )} d x

ln ∣ 2 x - 1 ∣ - 6 ln ∣ 2 x - 3 ∣ + 5 ln ∣ 2 x - 5 ∣ + C

Solution steps

\int \frac{32 x}{( 2 x - 1 ) ( 4 x ^{2} - 16 x + 15 )} d x

Take the constant out:

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

= 32 \cdot \int \frac{x}{( 2 x - 1 ) ( 4 x ^{2} - 16 x + 15 )} d x

Take the partial fraction of

\frac{x}{( 2 x - 1 ) ( 4 x ^{2} - 16 x + 15 )} : \frac{1}{16 ( 2 x - 1 )} - \frac{3}{8 ( 2 x - 3 )} + \frac{5}{16 ( 2 x - 5 )}

= 32 \cdot \int \frac{1}{16 ( 2 x - 1 )} - \frac{3}{8 ( 2 x - 3 )} + \frac{5}{16 ( 2 x - 5 )} 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

= 32 (\int \frac{1}{16 ( 2 x - 1 )} d x - \int \frac{3}{8 ( 2 x - 3 )} d x + \int \frac{5}{16 ( 2 x - 5 )} d x)

\int \frac{1}{16 ( 2 x - 1 )} d x = \frac{1}{32} ln ∣ 2 x - 1 ∣

\int \frac{3}{8 ( 2 x - 3 )} d x = \frac{3}{16} ln ∣ 2 x - 3 ∣

\int \frac{5}{16 ( 2 x - 5 )} d x = \frac{5}{32} ln ∣ 2 x - 5 ∣

= 32 (\frac{1}{32} ln ∣ 2 x - 1 ∣ - \frac{3}{16} ln ∣ 2 x - 3 ∣ + \frac{5}{32} ln ∣ 2 x - 5 ∣)

Simplify

32 (\frac{1}{32} ln ∣ 2 x - 1 ∣ - \frac{3}{16} ln ∣ 2 x - 3 ∣ + \frac{5}{32} ln ∣ 2 x - 5 ∣) : ln ∣ 2 x - 1 ∣ - 6 ln ∣ 2 x - 3 ∣ + 5 ln ∣ 2 x - 5 ∣

= ln ∣ 2 x - 1 ∣ - 6 ln ∣ 2 x - 3 ∣ + 5 ln ∣ 2 x - 5 ∣

Add a constant to the solution

= ln ∣ 2 x - 1 ∣ - 6 ln ∣ 2 x - 3 ∣ + 5 ln ∣ 2 x - 5 ∣ + C

\int \frac{3 x - 1}{x ^{2} - 2 x - 3} d x

\int \frac{1}{3 + 9 - x ^{2}} d x

\int 8 sin (- 4 x) cos^{3} (- 4 x) d x

\int x^{2} (ln (x))^{3} d x

\int \frac{1}{sin ( t ) - cos ( t )} d t

What is the integral of (32x)/((2x-1)(4x^2-16x+15)) ?
The integral of (32x)/((2x-1)(4x^2-16x+15)) is ln|2x-1|-6ln|2x-3|+5ln|2x-5|+C