What is the integral of 1/(xsqrt(1+4x)) ?

The integral of 1/(xsqrt(1+4x)) is -ln|sqrt(1+4x)+1|+ln|sqrt(1+4x)-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(1+4x))

\int \frac{1}{x 1 + 4 x} d x

- ln 1 + 4 x + 1 + ln 1 + 4 x - 1 + C

Solution steps

\int \frac{1}{x 1 + 4 x} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Factor

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

= 2 \cdot \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

= 2 (- \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}

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

Substitute back

u = 1 + 4 x

= 2 (- (\frac{ln 1 + 4 x + 1}{2} - \frac{ln 1 + 4 x - 1}{2}))

Simplify

2 (- (\frac{ln 1 + 4 x + 1}{2} - \frac{ln 1 + 4 x - 1}{2})) : - ln 1 + 4 x + 1 + ln 1 + 4 x - 1

= - ln 1 + 4 x + 1 + ln 1 + 4 x - 1

Add a constant to the solution

= - ln 1 + 4 x + 1 + ln 1 + 4 x - 1 + C

\int 5 x^{4} sinh (x^{5}) d x

\int arcsin (2 y) d y

\int \frac{10 x}{3 + 5 x ^{2}} d x

\int \frac{x}{x ^{4} - 2 x ^{2} + 10} d x

\int (\frac{5}{t} - 3 e^{t}) d t

What is the integral of 1/(xsqrt(1+4x)) ?
The integral of 1/(xsqrt(1+4x)) is -ln|sqrt(1+4x)+1|+ln|sqrt(1+4x)-1|+C