What is the integral of (-5)/(sqrt(x^2+16)) ?

The integral of (-5)/(sqrt(x^2+16)) is -5ln(1/4 |x+sqrt(16+x^2)|)+C

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integral of (-5)/(sqrt(x^2+16))

\int \frac{- 5}{x ^{2} + 16} d x

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

Solution steps

\int \frac{- 5}{x ^{2} + 16} d x

Take the constant out:

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

= - 5 \cdot \int \frac{1}{x ^{2} + 16} d x

Apply Trigonometric Substitution

: \int sec (u) d u

= - 5 \cdot \int sec (u) d u

Use the common integral:

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

= - 5 ln ∣ tan (u) + sec (u) ∣

Substitute back

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

= - 5 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x))

Simplify

- 5 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)) : - 5 ln (\frac{1}{4} x + 16 + x^{2})

= - 5 ln (\frac{1}{4} x + 16 + x^{2})

Add a constant to the solution

= - 5 ln (\frac{1}{4} x + 16 + x^{2}) + C

\int \frac{2 e ^{- \frac{6}{x ^{4}}}}{x ^{5}} d x

\frac{d ^{2} y}{d t ^{2}} - 8 \frac{d y}{d t} + 16 y = 0

f (x) = x^{3}, at x = 1

y = ln (18 x^{2} - 5 x + 11)

2^{x + 3}

What is the integral of (-5)/(sqrt(x^2+16)) ?
The integral of (-5)/(sqrt(x^2+16)) is -5ln(1/4 |x+sqrt(16+x^2)|)+C