What is the integral of 6/(5-12w+9w^2) ?

The integral of 6/(5-12w+9w^2) is 2arctan(3w-2)+C

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integral of 6/(5-12w+9w^2)

\int \frac{6}{5 - 12 w + 9 w ^{2}} d w

2 arctan (3 w - 2) + C

Solution steps

\int \frac{6}{5 - 12 w + 9 w ^{2}} d w

Take the constant out:

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

= 6 \cdot \int \frac{1}{5 - 12 w + 9 w ^{2}} d w

Complete the square

5 - 12 w + 9 w^{2} : 9 (w - \frac{2}{3})^{2} + 1

= 6 \cdot \int \frac{1}{9 ( w - \frac{2}{3} ) ^{2} + 1} d w

Apply u-substitution

: \int \frac{1}{9 u ^{2} + 1} d u

= 6 \cdot \int \frac{1}{9 u ^{2} + 1} d u

Apply Integral Substitution

= 6 \cdot \int \frac{1}{3 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 6 \cdot \frac{1}{3} arctan (v)

Substitute back

= 6 \cdot \frac{1}{3} arctan (3 (w - \frac{2}{3}))

Simplify

6 \cdot \frac{1}{3} arctan (3 (w - \frac{2}{3})) : 2 arctan (3 w - 2)

= 2 arctan (3 w - 2)

Add a constant to the solution

= 2 arctan (3 w - 2) + C

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What is the integral of 6/(5-12w+9w^2) ?
The integral of 6/(5-12w+9w^2) is 2arctan(3w-2)+C