What is the integral of 1/(5x^2-10x+6) ?

The integral of 1/(5x^2-10x+6) is 1/(sqrt(5))arctan(sqrt(5)(x-1))+C

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integral of 1/(5x^2-10x+6)

\int \frac{1}{5 x ^{2} - 10 x + 6} d x

\frac{1}{5} arctan (5 (x - 1)) + C

Solution steps

\int \frac{1}{5 x ^{2} - 10 x + 6} d x

Complete the square

5 x^{2} - 10 x + 6 : 5 (x - 1)^{2} + 1

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

Apply u-substitution

: \int \frac{1}{5 u ^{2} + 1} d u

= \int \frac{1}{5 u ^{2} + 1} d u

Apply Integral Substitution

= \int \frac{1}{5 ( v ^{2} + 1 )} d v

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{5} arctan (v)

Substitute back

= \frac{1}{5} arctan (5 (x - 1))

Add a constant to the solution

= \frac{1}{5} arctan (5 (x - 1)) + C

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What is the integral of 1/(5x^2-10x+6) ?
The integral of 1/(5x^2-10x+6) is 1/(sqrt(5))arctan(sqrt(5)(x-1))+C