What is the integral of (-4)/((sqrt(x^2+16))) ?

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

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integral of (-4)/((sqrt(x^2+16)))

\int \frac{- 4}{( x ^{2} + 16 )} d x

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

Solution steps

\int \frac{- 4}{x ^{2} + 16} d x

Take the constant out:

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

= - 4 \cdot \int \frac{1}{x ^{2} + 16} d x

Apply Trigonometric Substitution

: \int sec (u) d u

= - 4 \cdot \int sec (u) d u

Use the common integral:

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

= - 4 ln ∣ tan (u) + sec (u) ∣

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{( x ^{2} - 2 x + 5 ) ^{2}} d x

\int \frac{4 x - 3}{x + 1} d x

f (x) = - x^{4} + 18 x^{2} - 11

\frac{6}{s}

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