What is the integral of 1/(sqrt(25+9x^2)) ?

The integral of 1/(sqrt(25+9x^2)) is 1/3 ln((|3x+sqrt(25+9x^2)|)/5)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(25+9x^2))

\int \frac{1}{25 + 9 x ^{2}} d x

\frac{1}{3} ln (\frac{3 x + 25 + 9 x ^{2}}{5}) + C

Solution steps

\int \frac{1}{25 + 9 x ^{2}} d x

Apply Trigonometric Substitution

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

= \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

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

Use the common integral:

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

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

Substitute back

u = arctan (\frac{3}{5} x)

= \frac{1}{3} ln tan (arctan (\frac{3}{5} x)) + sec (arctan (\frac{3}{5} x))

Simplify

\frac{1}{3} ln tan (arctan (\frac{3}{5} x)) + sec (arctan (\frac{3}{5} x)) : \frac{1}{3} ln (\frac{3 x + 25 + 9 x ^{2}}{5})

= \frac{1}{3} ln (\frac{3 x + 25 + 9 x ^{2}}{5})

Add a constant to the solution

= \frac{1}{3} ln (\frac{3 x + 25 + 9 x ^{2}}{5}) + C

\int e^{- s t} sin (4 t) d t

\int \frac{2}{1 + tan ( x )} d x

\int_{4}^{8} t^{4} ln (3 t) d t

f (x) = x (2 - x)^{2}

\int t^{2} e^{5 t} d t

What is the integral of 1/(sqrt(25+9x^2)) ?
The integral of 1/(sqrt(25+9x^2)) is 1/3 ln((|3x+sqrt(25+9x^2)|)/5)+C