What is the integral of 5/(sqrt(9-16x^2)) ?

The integral of 5/(sqrt(9-16x^2)) is 5/4 arcsin(4/3 x)+C

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integral of 5/(sqrt(9-16x^2))

\int \frac{5}{9 - 16 x ^{2}} d x

\frac{5}{4} arcsin (\frac{4}{3} x) + C

Solution steps

\int \frac{5}{9 - 16 x ^{2}} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{9 - 16 x ^{2}} d x

Apply Trigonometric Substitution

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

= 5 \cdot \int \frac{1}{4} d u

Take the constant out:

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

= 5 \cdot \frac{1}{4} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 5 \cdot \frac{1}{4} \cdot 1 \cdot u

Substitute back

u = arcsin (\frac{4}{3} x)

= 5 \cdot \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{3} x)

Simplify

5 \cdot \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{3} x) : \frac{5}{4} arcsin (\frac{4}{3} x)

= \frac{5}{4} arcsin (\frac{4}{3} x)

Add a constant to the solution

= \frac{5}{4} arcsin (\frac{4}{3} x) + C

\frac{d y}{d t} = - k y

x \to - 1 lim (- x^{2} + 4 x - 5)

\frac{d}{d x} (\frac{2 x ^{2}}{( x ^{2} + x ) ^{3}})

f (x) = 4 x^{2} - 2 x + 4, at x = 0

5 - 9 + 8 x^{2}

What is the integral of 5/(sqrt(9-16x^2)) ?
The integral of 5/(sqrt(9-16x^2)) is 5/4 arcsin(4/3 x)+C