What is the integral of arcsin(4/3 x+5) ?

The integral of arcsin(4/3 x+5) is 1/4 (4xarcsin(4/3 x+5)+15arcsin(4/3 x+5)+sqrt(-16x^2-120x-216))+C

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integral of arcsin(4/3 x+5)

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

\frac{1}{4} (4 x arcsin (\frac{4}{3} x + 5) + 15 arcsin (\frac{4}{3} x + 5) + - 16 x^{2} - 120 x - 216) + C

Solution steps

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

Apply u-substitution

: \int arcsin (u) \frac{3}{4} d u

= \int arcsin (u) \frac{3}{4} d u

Take the constant out:

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

= \frac{3}{4} \cdot \int arcsin (u) d u

Apply Integration By Parts

: u arcsin (u) - \int \frac{u}{1 - u ^{2}} d u

= \frac{3}{4} (u arcsin (u) - \int \frac{u}{1 - u ^{2}} d u)

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

= \frac{3}{4} (u arcsin (u) - (- 1 - u^{2}))

Substitute back

u = \frac{4}{3} x + 5

= \frac{3}{4} (\frac{4}{3} x + 5) arcsin (\frac{4}{3} x + 5) - - 1 - (\frac{4}{3} x + 5)^{2}

Simplify

\frac{3}{4} (\frac{4}{3} x + 5) arcsin (\frac{4}{3} x + 5) - - 1 - (\frac{4}{3} x + 5)^{2} : \frac{1}{4} (4 x arcsin (\frac{4}{3} x + 5) + 15 arcsin (\frac{4}{3} x + 5) + - 16 x^{2} - 120 x - 216)

= \frac{1}{4} (4 x arcsin (\frac{4}{3} x + 5) + 15 arcsin (\frac{4}{3} x + 5) + - 16 x^{2} - 120 x - 216)

Add a constant to the solution

= \frac{1}{4} (4 x arcsin (\frac{4}{3} x + 5) + 15 arcsin (\frac{4}{3} x + 5) + - 16 x^{2} - 120 x - 216) + C

(x^{3} + 2)^{5}, at x = - 1

\frac{\partial}{\partial x} (56 y^{4} x^{6} - 42 y^{3} x^{5})

(x - 1)^{2} (x - 2)^{3}

n = 1 \sum \infty 8 ((\frac{1}{5})^{n})

x \to 0 lim (\frac{6}{x} - \frac{6}{x ^{2} - x})

What is the integral of arcsin(4/3 x+5) ?
The integral of arcsin(4/3 x+5) is 1/4 (4xarcsin(4/3 x+5)+15arcsin(4/3 x+5)+sqrt(-16x^2-120x-216))+C