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

The integral of (2x-2)/((x-1)^3+4(x-1)) is arctan(1/2 (x-1))+C

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

\int \frac{2 x - 2}{( x - 1 ) ^{3} + 4 ( x - 1 )} d x

arctan (\frac{1}{2} (x - 1)) + C

Solution steps

\int \frac{2 x - 2}{( x - 1 ) ^{3} + 4 ( x - 1 )} d x

Take the partial fraction of

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

= \int \frac{2}{x ^{2} - 2 x + 5} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{x ^{2} - 2 x + 5} d x

Complete the square

x^{2} - 2 x + 5 : (x - 1)^{2} + 4

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

Apply u-substitution

= 2 \cdot \int \frac{1}{u ^{2} + 4} d u

Apply Integral Substitution

= 2 \cdot \int \frac{1}{2 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 2 \cdot \frac{1}{2} arctan (v)

Substitute back

= 2 \cdot \frac{1}{2} arctan (\frac{x - 1}{2})

Simplify

2 \cdot \frac{1}{2} arctan (\frac{x - 1}{2}) : arctan (\frac{1}{2} (x - 1))

= arctan (\frac{1}{2} (x - 1))

Add a constant to the solution

= arctan (\frac{1}{2} (x - 1)) + C

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What is the integral of (2x-2)/((x-1)^3+4(x-1)) ?
The integral of (2x-2)/((x-1)^3+4(x-1)) is arctan(1/2 (x-1))+C