What is the integral of 1/(xsqrt(x^2+49)) ?

The integral of 1/(xsqrt(x^2+49)) is 1/7 ln|tan((arctan(1/7 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(x^2+49))

\int \frac{1}{x x ^{2} + 49} d x

\frac{1}{7} ln tan (\frac{arctan ( \frac{1}{7} x )}{2}) + C

Solution steps

\int \frac{1}{x x ^{2} + 49} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{7 tan ( u )} d u

= \int \frac{sec ( u )}{7 tan ( u )} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int \frac{sec ( u )}{tan ( u )} d u

Apply u-substitution

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

= \frac{1}{7} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{7} ln ∣ v ∣

Substitute back

= \frac{1}{7} ln tan (\frac{arctan ( \frac{1}{7} x )}{2})

Add a constant to the solution

= \frac{1}{7} ln tan (\frac{arctan ( \frac{1}{7} x )}{2}) + C

\int \frac{x ^{2} - 4 x + 6}{x - 1} d x

\int \frac{1 + cos ^{2} ( x )}{sin ^{2} ( x )} d x

\int \frac{1}{30} e^{- \frac{x}{30}} d x

\int 4 x cos (2 x^{2}) d x

\int \frac{1}{x ^{2} - 10 x + 34} d x

What is the integral of 1/(xsqrt(x^2+49)) ?
The integral of 1/(xsqrt(x^2+49)) is 1/7 ln|tan((arctan(1/7 x))/2)|+C