What is the integral of 6/(sqrt(x^2+16)) ?

The integral of 6/(sqrt(x^2+16)) is 6ln(1/4 |x+sqrt(16+x^2)|)+C

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integral of 6/(sqrt(x^2+16))

\int \frac{6}{x ^{2} + 16} d x

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

Solution steps

\int \frac{6}{x ^{2} + 16} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{x ^{2} + 16} d x

Apply Trigonometric Substitution

: \int sec (u) d u

= 6 \cdot \int sec (u) d u

Use the common integral:

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

= 6 ln ∣ tan (u) + sec (u) ∣

Substitute back

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

= 6 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x))

Simplify

6 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)) : 6 ln (\frac{1}{4} x + 16 + x^{2})

= 6 ln (\frac{1}{4} x + 16 + x^{2})

Add a constant to the solution

= 6 ln (\frac{1}{4} x + 16 + x^{2}) + C

y^{^{''}} - 8 y^{^{'}} + 15 y = 0, y (0) = 1, y^{^{'}} (0) = 4

\frac{d y}{d t} + y = 2, y (0) = 0

\int cos^{4} (θ) sin (θ) d θ

\frac{d}{d x} (ln (3 x^{2} - 2 x))

\frac{d}{d x} (ln ((x^{2} + 1)^{3}))

What is the integral of 6/(sqrt(x^2+16)) ?
The integral of 6/(sqrt(x^2+16)) is 6ln(1/4 |x+sqrt(16+x^2)|)+C