What is the integral of (-8)/(sqrt(x^2+16)) ?

The integral of (-8)/(sqrt(x^2+16)) is -8ln(1/4 |x+sqrt(16+x^2)|)+C

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integral of (-8)/(sqrt(x^2+16))

\int \frac{- 8}{x ^{2} + 16} d x

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

Solution steps

\int \frac{- 8}{x ^{2} + 16} d x

Take the constant out:

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

= - 8 \cdot \int \frac{1}{x ^{2} + 16} d x

Apply Trigonometric Substitution

= - 8 \cdot \int sec (u) d u

Use the common integral:

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

= - 8 ln ∣ tan (u) + sec (u) ∣

Substitute back

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

= - 8 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x))

Simplify

- 8 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)) : - 8 ln (\frac{1}{4} x + 16 + x^{2})

= - 8 ln (\frac{1}{4} x + 16 + x^{2})

Add a constant to the solution

= - 8 ln (\frac{1}{4} x + 16 + x^{2}) + C

\frac{d}{d x} (arctan (sin (7 x)))

\int o d o

2 y y^{^{'}} + 2 = y^{2} + 2 x, y (0) = 8

\frac{d}{d x} (x arcsec (x^{3}))

in v erse l a pl a ce \frac{( 6 + 6 s )}{2 s ^{2} + 1}

What is the integral of (-8)/(sqrt(x^2+16)) ?
The integral of (-8)/(sqrt(x^2+16)) is -8ln(1/4 |x+sqrt(16+x^2)|)+C