What is the integral of 1/(sqrt(-t^2+6t-8)) ?

The integral of 1/(sqrt(-t^2+6t-8)) is arcsin(t-3)+C

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integral of 1/(sqrt(-t^2+6t-8))

\int \frac{1}{- t ^{2} + 6 t - 8} d t

arcsin (t - 3) + C

Solution steps

\int \frac{1}{- t ^{2} + 6 t - 8} d t

Complete the square

- t^{2} + 6 t - 8 : - (t - 3)^{2} + 1

= \int \frac{1}{- ( t - 3 ) ^{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 - 3

= arcsin (t - 3)

Add a constant to the solution

= arcsin (t - 3) + C

\frac{d}{d x} (0.006 x^{3} + 0.02 x^{2} + 0.5 x)

\frac{d}{d y} (y cos (x) y)

3 x^{2} - 9

y^{^{'}} = \frac{e ^{x}}{y ^{2}}

\frac{\partial}{\partial x} (2 y + z)

What is the integral of 1/(sqrt(-t^2+6t-8)) ?
The integral of 1/(sqrt(-t^2+6t-8)) is arcsin(t-3)+C