What is the integral of 3/(sqrt(1+9x^2)) ?

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

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integral of 3/(sqrt(1+9x^2))

\int \frac{3}{1 + 9 x ^{2}} d x

ln 3 x + 1 + 9 x^{2} + C

Solution steps

\int \frac{3}{1 + 9 x ^{2}} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{1 + 9 x ^{2}} d x

Apply Trigonometric Substitution

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

= 3 \cdot \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

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

Use the common integral:

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

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

Substitute back

u = arctan (3 x)

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

Simplify

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

= ln 3 x + 1 + 9 x^{2}

Add a constant to the solution

= ln 3 x + 1 + 9 x^{2} + C

\int \frac{y ^{2} - 4}{y} d y

\frac{d}{d x} (\frac{x ^{5} - 8 x ^{3} + x + 6}{x})

x \to 0 lim (2 x^{3} - 4 x + 2)

\int (\frac{2}{x} + 3 sin (x) - 4 e^{x}) d x

y = 1 + 3 x, 1 \leq x \leq 9

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