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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{4} d u

= \int \frac{sec ( u )}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{4} ln ∣ tan (u) + sec (u) ∣

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

(x^{2} + x y) d x - x^{2} d y = 0

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y = x^{4} + 8 x^{2} - x

\int 4 \cdot sin (2 x) - 3 \cdot cos (3 x) d x

\int 15 x^{17} e^{x^{9}} d x

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