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

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

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

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

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

Solution steps

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

f (x) = 8 x

\int \frac{1}{9 x ^{2} - 1} d x

(x + y) d x = (y - x) d y

\int \frac{21}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

(- 10 - 6) (- 10 - 5)

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