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

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

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

\int \frac{1}{- x ^{2} + 4 x - 1} d x

arcsin (\frac{1}{3} (x - 2)) + C

Solution steps

\int \frac{1}{- x ^{2} + 4 x - 1} d x

Complete the square

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

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

Apply u-substitution

: \int \frac{1}{- u ^{2} + 3} d u

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

Apply Trigonometric Substitution

: \int 1 d v

= \int 1 d v

Integral of a constant:

\int a d x = a x

= 1 \cdot v

Substitute back

= 1 \cdot arcsin (\frac{1}{3} (x - 2))

Simplify

= arcsin (\frac{1}{3} (x - 2))

Add a constant to the solution

= arcsin (\frac{1}{3} (x - 2)) + C

y = x^{2} - 12 x + 4

\int e^{t} sin (3 t) d t

2 tan^{2} (4 x)

\frac{\partial}{\partial x} (x^{2} y)

y = lo g_{3} (sin (x^{2}))

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