What is the integral of 1/(xsqrt(169x^2-1)) ?

The integral of 1/(xsqrt(169x^2-1)) is arctan(sqrt(169x^2-1))+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(169x^2-1))

\int \frac{1}{x 169 x ^{2} - 1} d x

arctan (169 x^{2} - 1) + C

Solution steps

\int \frac{1}{x 169 x ^{2} - 1} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 1} d u

= \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= arctan (u)

Substitute back

u = 169 x^{2} - 1

= arctan (169 x^{2} - 1)

Add a constant to the solution

= arctan (169 x^{2} - 1) + C

\int e^{- 0.6 t} d t

\int \frac{e ^{7 x}}{e ^{7 x} + 6} d x

\int (\frac{x}{16 - x ^{2}}) d x

\int \frac{3 x}{( x + 1 ) ^{3}} d x

\int t e^{t - 1} d t

What is the integral of 1/(xsqrt(169x^2-1)) ?
The integral of 1/(xsqrt(169x^2-1)) is arctan(sqrt(169x^2-1))+C