What is the integral of 1/(sqrt(9+4x^2)) ?

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

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integral of 1/(sqrt(9+4x^2))

\int \frac{1}{9 + 4 x ^{2}} d x

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

Solution steps

\int \frac{1}{9 + 4 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{2} d u

= \int \frac{sec ( u )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{2}{3} x)

= \frac{1}{2} ln tan (arctan (\frac{2}{3} x)) + sec (arctan (\frac{2}{3} x))

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{9}{2 x + 1})

\int_{1 - 3 x}^{1} \frac{u ^{3}}{1 + u ^{2}} d u

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

n \to \infty lim (5)

\int - x^{2} + 6 x - 5 d x

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