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

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

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

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

arcsin (2 t - 3) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

= 2 \cdot \int \frac{1}{2} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 2 \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

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

Simplify

2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (2 (t - \frac{3}{2})) : arcsin (2 t - 3)

= arcsin (2 t - 3)

Add a constant to the solution

= arcsin (2 t - 3) + C

\int e^{\frac{x ^{2}}{4}} d x

x y \frac{d y}{d x} = 7

\int_{0}^{1} sin (π t + \frac{π}{4}) d t

\int 3 w d w

y^{^{'''}} + 16 y^{^{'}} = 0

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