What is the integral of 1/(sqrt(9x^2-6x+5)) ?

The integral of 1/(sqrt(9x^2-6x+5)) is 1/3 ln((|3x-1+sqrt(9x^2-6x+5)|)/2)+C

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integral of 1/(sqrt(9x^2-6x+5))

\int \frac{1}{9 x ^{2} - 6 x + 5} d x

\frac{1}{3} ln (\frac{3 x - 1 + 9 x ^{2} - 6 x + 5}{2}) + C

Solution steps

\int \frac{1}{9 x ^{2} - 6 x + 5} d x

Complete the square

9 x^{2} - 6 x + 5 : 9 (x - \frac{1}{3})^{2} + 4

= \int \frac{1}{9 ( x - \frac{1}{3} ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{1}{9 u ^{2} + 4} d u

= \int \frac{1}{9 u ^{2} + 4} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{3} d v

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

\frac{1}{3} ln tan (arctan (\frac{3}{2} (x - \frac{1}{3}))) + sec (arctan (\frac{3}{2} (x - \frac{1}{3}))) : \frac{1}{3} ln (\frac{3 x - 1 + 9 x ^{2} - 6 x + 5}{2})

= \frac{1}{3} ln (\frac{3 x - 1 + 9 x ^{2} - 6 x + 5}{2})

Add a constant to the solution

= \frac{1}{3} ln (\frac{3 x - 1 + 9 x ^{2} - 6 x + 5}{2}) + C

\int \frac{9 - 162 x}{4 - 81 x ^{2}} d x

\int_{1}^{8} \frac{x}{x ^{2} + 4 x + 20} d x

\frac{1}{x ^{4}}

\int_{0}^{4} (4 e^{- 0.25 x} - 6 x) d x

\frac{x}{lo g _{e} ( x )}

What is the integral of 1/(sqrt(9x^2-6x+5)) ?
The integral of 1/(sqrt(9x^2-6x+5)) is 1/3 ln((|3x-1+sqrt(9x^2-6x+5)|)/2)+C