What is the integral of 1/(sqrt(16x^2+25)) ?

The integral of 1/(sqrt(16x^2+25)) is 1/4 ln((|4x+sqrt(25+16x^2)|)/5)+C

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integral of 1/(sqrt(16x^2+25))

\int \frac{1}{16 x ^{2} + 25} d x

\frac{1}{4} ln (\frac{4 x + 25 + 16 x ^{2}}{5}) + C

Solution steps

\int \frac{1}{16 x ^{2} + 25} 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}{5} x)

= \frac{1}{4} ln tan (arctan (\frac{4}{5} x)) + sec (arctan (\frac{4}{5} x))

Simplify

\frac{1}{4} ln tan (arctan (\frac{4}{5} x)) + sec (arctan (\frac{4}{5} x)) : \frac{1}{4} ln (\frac{4 x + 25 + 16 x ^{2}}{5})

= \frac{1}{4} ln (\frac{4 x + 25 + 16 x ^{2}}{5})

Add a constant to the solution

= \frac{1}{4} ln (\frac{4 x + 25 + 16 x ^{2}}{5}) + C

\int 15 - 2 x - x^{2} d x

\frac{d y}{d t} + k (340 - y) = 0

e

\int - 4 x^{2} (4 x^{3} - 1)^{7} d x

f (x) = x^{2 x}

What is the integral of 1/(sqrt(16x^2+25)) ?
The integral of 1/(sqrt(16x^2+25)) is 1/4 ln((|4x+sqrt(25+16x^2)|)/5)+C