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

The integral of 1/(xsqrt(49x^2+1)) is -1/2 ln|sqrt(49x^2+1)+1|+1/2 ln|sqrt(49x^2+1)-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(49x^2+1))

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

- \frac{1}{2} ln 49 x^{2} + 1 + 1 + \frac{1}{2} ln 49 x^{2} + 1 - 1 + C

Solution steps

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

Apply u-substitution

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

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

Factor

u^{2} - 1 : - (- u^{2} + 1)

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

Take the constant out:

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

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

Use the common integral:

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

= - (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = 49 x^{2} + 1

= - (\frac{ln 49 x ^{2} + 1 + 1}{2} - \frac{ln 49 x ^{2} + 1 - 1}{2})

Simplify

- (\frac{ln 49 x ^{2} + 1 + 1}{2} - \frac{ln 49 x ^{2} + 1 - 1}{2}) : - \frac{1}{2} ln 49 x^{2} + 1 + 1 + \frac{1}{2} ln 49 x^{2} + 1 - 1

= - \frac{1}{2} ln 49 x^{2} + 1 + 1 + \frac{1}{2} ln 49 x^{2} + 1 - 1

Add a constant to the solution

= - \frac{1}{2} ln 49 x^{2} + 1 + 1 + \frac{1}{2} ln 49 x^{2} + 1 - 1 + C

\int sec^{7} (5 x + 4) d x

\frac{x}{1 + x ^{2}}

x \to 4 + lim (\frac{7}{( x - 4 ) ^{2}})

\frac{d}{d y} ((cos (x) + sin (y))^{y s i n (2 x - y)})

2 x ln (x + 1)

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