What is the integral of 1/(sqrt(-t^2+8t-15)) ?

The integral of 1/(sqrt(-t^2+8t-15)) is arcsin(t-4)+C

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integral of 1/(sqrt(-t^2+8t-15))

\int \frac{1}{- t ^{2} + 8 t - 15} d t

arcsin (t - 4) + C

Solution steps

\int \frac{1}{- t ^{2} + 8 t - 15} d t

Complete the square

- t^{2} + 8 t - 15 : - (t - 4)^{2} + 1

= \int \frac{1}{- ( t - 4 ) ^{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 - 4

= arcsin (t - 4)

Add a constant to the solution

= arcsin (t - 4) + C

x + e^{x} - \frac{1}{x}

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\int \frac{9}{6 + 9 x} d x

What is the integral of 1/(sqrt(-t^2+8t-15)) ?
The integral of 1/(sqrt(-t^2+8t-15)) is arcsin(t-4)+C