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

The integral of (-9)/(sqrt(x^2+16)) is -9ln(1/4 |x+sqrt(16+x^2)|)+C

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

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

- 9 ln (\frac{1}{4} x + 16 + x^{2}) + C

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int sec (u) d u

= - 9 \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= - 9 ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{1}{4} x)

= - 9 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x))

Simplify

- 9 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)) : - 9 ln (\frac{1}{4} x + 16 + x^{2})

= - 9 ln (\frac{1}{4} x + 16 + x^{2})

Add a constant to the solution

= - 9 ln (\frac{1}{4} x + 16 + x^{2}) + C

\frac{d ^{2}}{d x ^{2}} ((3 x^{2} + 6 x) sin (x))

\frac{d}{d x} (cos (2 x - 1) tan (1 - 2 x))

e^{x}, x e^{x^{2}}, [0, 1]

f (x) = \frac{1}{( 3 - x ) ^{2}}

What is the integral of (-9)/(sqrt(x^2+16)) ?
The integral of (-9)/(sqrt(x^2+16)) is -9ln(1/4 |x+sqrt(16+x^2)|)+C