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

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

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

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

\frac{1}{2} arcsin (\frac{4}{3} t + \frac{1}{3}) + C

Solution steps

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

Complete the square

1 - t - 2 t^{2} : - 2 (t + \frac{1}{4})^{2} + \frac{9}{8}

= \int \frac{1}{- 2 ( t + \frac{1}{4} ) ^{2} + \frac{9}{8}} d t

Apply u-substitution

: \int \frac{2 2}{- 16 u ^{2} + 9} d u

= \int \frac{2 2}{- 16 u ^{2} + 9} d u

Take the constant out:

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

= 22 \cdot \int \frac{1}{- 16 u ^{2} + 9} d u

Apply Trigonometric Substitution

: \int \frac{1}{4} d v

= 22 \cdot \int \frac{1}{4} d v

Take the constant out:

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

= 22 \frac{1}{4} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 22 \frac{1}{4} \cdot 1 \cdot v

Substitute back

= 22 \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{3} (t + \frac{1}{4}))

Simplify

22 \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{3} (t + \frac{1}{4})) : \frac{1}{2} arcsin (\frac{4}{3} t + \frac{1}{3})

= \frac{1}{2} arcsin (\frac{4}{3} t + \frac{1}{3})

Add a constant to the solution

= \frac{1}{2} arcsin (\frac{4}{3} t + \frac{1}{3}) + C

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