What is the integral of 1/(sqrt(9x^2+16)) ?

The integral of 1/(sqrt(9x^2+16)) is 1/3 ln((|3x+sqrt(16+9x^2)|)/4)+C

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integral of 1/(sqrt(9x^2+16))

\int \frac{1}{9 x ^{2} + 16} d x

\frac{1}{3} ln (\frac{3 x + 16 + 9 x ^{2}}{4}) + C

Solution steps

\int \frac{1}{9 x ^{2} + 16} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{3} d u

= \int \frac{sec ( u )}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{3} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{3}{4} x)

= \frac{1}{3} ln tan (arctan (\frac{3}{4} x)) + sec (arctan (\frac{3}{4} x))

Simplify

\frac{1}{3} ln tan (arctan (\frac{3}{4} x)) + sec (arctan (\frac{3}{4} x)) : \frac{1}{3} ln (\frac{3 x + 16 + 9 x ^{2}}{4})

= \frac{1}{3} ln (\frac{3 x + 16 + 9 x ^{2}}{4})

Add a constant to the solution

= \frac{1}{3} ln (\frac{3 x + 16 + 9 x ^{2}}{4}) + C

h \to 0 + lim (\frac{∣ h ∣}{h} + h^{2})

e^{f (x)}

2 x (1 - x)

\frac{d y}{d x} = \frac{8 x}{3 y}

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

What is the integral of 1/(sqrt(9x^2+16)) ?
The integral of 1/(sqrt(9x^2+16)) is 1/3 ln((|3x+sqrt(16+9x^2)|)/4)+C