What is the integral of 1/(3sqrt(2x^2-20x+68)) ?

The integral of 1/(3sqrt(2x^2-20x+68)) is 1/(3sqrt(2))ln((|x-5+sqrt(x^2-10x+34)|)/3)+C

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integral of 1/(3sqrt(2x^2-20x+68))

\int \frac{1}{3 2 x ^{2} - 20 x + 68} d x

\frac{1}{3 2} ln (\frac{x - 5 + x ^{2} - 10 x + 34}{3}) + C

Solution steps

\int \frac{1}{3 2 x ^{2} - 20 x + 68} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{2 x ^{2} - 20 x + 68} d x

Complete the square

2 x^{2} - 20 x + 68 : 2 (x - 5)^{2} + 18

= \frac{1}{3} \cdot \int \frac{1}{2 ( x - 5 ) ^{2} + 18} d x

Apply u-substitution

: \int \frac{1}{2 u ^{2} + 18} d u

= \frac{1}{3} \cdot \int \frac{1}{2 u ^{2} + 18} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{2} d v

= \frac{1}{3} \cdot \int \frac{sec ( v )}{2} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{2} \cdot \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= \frac{1}{3} \cdot \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{3} \cdot \frac{1}{2} ln tan (arctan (\frac{2}{3 2} (x - 5))) + sec (arctan (\frac{2}{3 2} (x - 5)))

Simplify

\frac{1}{3} \cdot \frac{1}{2} ln tan (arctan (\frac{2}{3 2} (x - 5))) + sec (arctan (\frac{2}{3 2} (x - 5))) : \frac{1}{3 2} ln (\frac{x - 5 + x ^{2} - 10 x + 34}{3})

= \frac{1}{3 2} ln (\frac{x - 5 + x ^{2} - 10 x + 34}{3})

Add a constant to the solution

= \frac{1}{3 2} ln (\frac{x - 5 + x ^{2} - 10 x + 34}{3}) + C

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What is the integral of 1/(3sqrt(2x^2-20x+68)) ?
The integral of 1/(3sqrt(2x^2-20x+68)) is 1/(3sqrt(2))ln((|x-5+sqrt(x^2-10x+34)|)/3)+C