What is the integral of 1/(xsqrt(16x^2+3)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(xsqrt(16x^2+3))

\int \frac{1}{x 16 x ^{2} + 3} d x

\frac{1}{3} ln tan \frac{arctan ( \frac{4}{3} x )}{2} + C

Solution steps

\int \frac{1}{x 16 x ^{2} + 3} d x

Apply Trigonometric Substitution

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

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

Apply u-substitution

: \int \frac{1}{v} d v

= \frac{1}{3} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{3} ln ∣ v ∣

Substitute back

= \frac{1}{3} ln tan \frac{arctan ( \frac{4}{3} x )}{2}

Add a constant to the solution

= \frac{1}{3} ln tan \frac{arctan ( \frac{4}{3} x )}{2} + C

\int e^{x} + 2 x d x

\int \frac{1}{θ ^{2}} csc (\frac{1}{θ}) d θ

\int e^{\frac{1}{2} \cdot x} d x

\int sin (2 x) ln (sin (x)) d x

\int \frac{a}{v ^{2}} d v

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