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 10 x^{5} - 10 x^{2} + 7 d x

\frac{d y}{d x} = \frac{x}{25 y}

\int_{0}^{1} 1 + x^{3} d x

x \to 1 lim (e^{x} (x^{3} - 4))

(5) (0.5)

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