What is the integral of 1/(sqrt(2-3x-4x^2)) ?

The integral of 1/(sqrt(2-3x-4x^2)) is 1/2 arcsin(1/(sqrt(41))(8x+3))+C

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integral of 1/(sqrt(2-3x-4x^2))

\int \frac{1}{2 - 3 x - 4 x ^{2}} d x

\frac{1}{2} arcsin (\frac{1}{41} (8 x + 3)) + C

Solution steps

\int \frac{1}{2 - 3 x - 4 x ^{2}} d x

Complete the square

2 - 3 x - 4 x^{2} : - 4 (x + \frac{3}{8})^{2} + \frac{41}{16}

= \int \frac{1}{- 4 ( x + \frac{3}{8} ) ^{2} + \frac{41}{16}} d x

Apply u-substitution

: \int \frac{4}{- 64 u ^{2} + 41} d u

= \int \frac{4}{- 64 u ^{2} + 41} d u

Take the constant out:

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

= 4 \cdot \int \frac{1}{- 64 u ^{2} + 41} d u

Apply Trigonometric Substitution

: \int \frac{1}{8} d v

= 4 \cdot \int \frac{1}{8} d v

Take the constant out:

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

= 4 \cdot \frac{1}{8} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 4 \cdot \frac{1}{8} \cdot 1 \cdot v

Substitute back

= 4 \cdot \frac{1}{8} \cdot 1 \cdot arcsin (\frac{8}{41} (x + \frac{3}{8}))

Simplify

4 \cdot \frac{1}{8} \cdot 1 \cdot arcsin (\frac{8}{41} (x + \frac{3}{8})) : \frac{1}{2} arcsin (\frac{1}{41} (8 x + 3))

= \frac{1}{2} arcsin (\frac{1}{41} (8 x + 3))

Add a constant to the solution

= \frac{1}{2} arcsin (\frac{1}{41} (8 x + 3)) + C

f (x) = e^{2 x} - e^{- 2 x}

f^{^{'}} (x) = 4 sin^{2} (x)

h \to 0 - lim (\frac{1}{h ^{2} + 1})

\int e^{- \frac{x}{11}} d x

x (\frac{d y}{d x}) = y + x e^{\frac{y}{x}}

What is the integral of 1/(sqrt(2-3x-4x^2)) ?
The integral of 1/(sqrt(2-3x-4x^2)) is 1/2 arcsin(1/(sqrt(41))(8x+3))+C