What is the integral of 1/((x-1)sqrt(4x^2-8x+3)) ?

The integral of 1/((x-1)sqrt(4x^2-8x+3)) is arctan(sqrt(4x^2-8x+3))+C

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integral of 1/((x-1)sqrt(4x^2-8x+3))

\int \frac{1}{( x - 1 ) 4 x ^{2} - 8 x + 3} d x

arctan (4 x^{2} - 8 x + 3) + C

Solution steps

\int \frac{1}{( x - 1 ) 4 x ^{2} - 8 x + 3} d x

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{1}{v ^{2} + 1} d v

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

Use the common integral:

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

= arctan (v)

Substitute back

= arctan (4 (x - 1)^{2} - 1)

Expand

4 (x - 1)^{2} - 1 : 4 x^{2} - 8 x + 3

= arctan (4 x^{2} - 8 x + 3)

Add a constant to the solution

= arctan (4 x^{2} - 8 x + 3) + C

\int x + \frac{2}{sin ^{2} ( x )} d x

y^{^{''}} - y = 0, y (0) = 0, y^{^{'}} (0) = 1

\frac{d}{d x} (8 cos^{4} (x))

x \to - 1 lim (f (x))

1. 1^{x} + 9^{x} d x

What is the integral of 1/((x-1)sqrt(4x^2-8x+3)) ?
The integral of 1/((x-1)sqrt(4x^2-8x+3)) is arctan(sqrt(4x^2-8x+3))+C