What is the integral of (6+x)/(sqrt(36-x^2)) ?

The integral of (6+x)/(sqrt(36-x^2)) is -sqrt(36-x^2)+6arcsin(1/6 x)+C

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integral of (6+x)/(sqrt(36-x^2))

\int \frac{6 + x}{36 - x ^{2}} d x

- 36 - x^{2} + 6 arcsin (\frac{1}{6} x) + C

Solution steps

\int \frac{6 + x}{36 - x ^{2}} d x

Apply Trigonometric Substitution

: \int 6 (sin (u) + 1) d u

= \int 6 (sin (u) + 1) d u

Take the constant out:

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

= 6 \cdot \int sin (u) + 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 6 (\int sin (u) d u + \int 1 d u)

\int sin (u) d u = - cos (u)

\int 1 d u = u

= 6 (- cos (u) + u)

Substitute back

u = arcsin (\frac{1}{6} x)

= 6 (- cos (arcsin (\frac{1}{6} x)) + arcsin (\frac{1}{6} x))

Simplify

6 (- cos (arcsin (\frac{1}{6} x)) + arcsin (\frac{1}{6} x)) : - 36 - x^{2} + 6 arcsin (\frac{1}{6} x)

= - 36 - x^{2} + 6 arcsin (\frac{1}{6} x)

Add a constant to the solution

= - 36 - x^{2} + 6 arcsin (\frac{1}{6} x) + C

\frac{d y}{d x}, 7 x y + 1 = 0

\frac{d}{d x} (9^{x} ln (x))

5 t^{2} - 6 cos (3 t) + 8 e^{- t}

x \to \infty lim (x + \frac{2}{x})

x \to 4 lim (5 x)

What is the integral of (6+x)/(sqrt(36-x^2)) ?
The integral of (6+x)/(sqrt(36-x^2)) is -sqrt(36-x^2)+6arcsin(1/6 x)+C