What is the integral of 1/(sqrt(-4t^2+12t+7)) ?

The integral of 1/(sqrt(-4t^2+12t+7)) is 1/2 arcsin(1/2 (t-3/2))+C

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integral of 1/(sqrt(-4t^2+12t+7))

\int \frac{1}{- 4 t ^{2} + 12 t + 7} d t

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

Solution steps

\int \frac{1}{- 4 t ^{2} + 12 t + 7} d t

Complete the square

- 4 t^{2} + 12 t + 7 : - 4 (t - \frac{3}{2})^{2} + 16

= \int \frac{1}{- 4 ( t - \frac{3}{2} ) ^{2} + 16} d t

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int 1 d v

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

Integral of a constant:

\int a d x = a x

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

Substitute back

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

Simplify

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

Add a constant to the solution

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

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What is the integral of 1/(sqrt(-4t^2+12t+7)) ?
The integral of 1/(sqrt(-4t^2+12t+7)) is 1/2 arcsin(1/2 (t-3/2))+C