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

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

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

\int \frac{1}{- t ^{2} + 4 t - 3} d t

arcsin (t - 2) + C

Solution steps

\int \frac{1}{- t ^{2} + 4 t - 3} d t

Complete the square

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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = arcsin (u)

= arcsin (u)

Substitute back

u = t - 2

= arcsin (t - 2)

Add a constant to the solution

= arcsin (t - 2) + C

ln (cos (3 x))

\frac{d}{d x} (x \cdot e^{- 2 x^{2}})

\int \frac{1}{5 - 3 x} d x

\frac{d}{d x} (- e^{- x} + \frac{1}{3})

\int_{2}^{\infty} - 6 x^{- 2} d x

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