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

\int t^{24} e^{- t^{25}} d t

x \to 0 lim ((x)^{3 x})

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

y = 9 x, y = x^{2} - 10

x^{4}, at x = - \frac{1}{3}

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