What is the integral of 1/(sqrt(-t^2-6t-5)) ?

The integral of 1/(sqrt(-t^2-6t-5)) is arcsin(1/2 (t+3))+C

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integral of 1/(sqrt(-t^2-6t-5))

\int \frac{1}{- t ^{2} - 6 t - 5} d t

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

Solution steps

\int \frac{1}{- t ^{2} - 6 t - 5} d t

Complete the square

- t^{2} - 6 t - 5 : - (t + 3)^{2} + 4

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

Apply u-substitution

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

= \int \frac{1}{- u ^{2} + 4} 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}{2} (t + 3))

Simplify

= arcsin (\frac{1}{2} (t + 3))

Add a constant to the solution

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

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

\int 85 - x d x

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

\int \frac{- 13 x + 3}{( - 4 x + 2 ) ^{2}} d x

\int (x^{2} - 2 x) e^{- x} d x

What is the integral of 1/(sqrt(-t^2-6t-5)) ?
The integral of 1/(sqrt(-t^2-6t-5)) is arcsin(1/2 (t+3))+C