What is the integral of 1/(x(1+x)^{1/2)} ?

The integral of 1/(x(1+x)^{1/2)} is -ln|(1+x)^{1/2}+1|+ln|(1+x)^{1/2}-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/(x(1+x)^{1/2)}

integral

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{1}{( 4 - cos ^{2} ( x ) )}

2 x^{- 2}

\frac{sin ( x ^{2} )}{x}

\frac{1}{x} \cdot cos^{2} (x)

1 + e^{- x} sin (4 x)

What is the integral of 1/(x(1+x)^{1/2)} ?
The integral of 1/(x(1+x)^{1/2)} is -ln|(1+x)^{1/2}+1|+ln|(1+x)^{1/2}-1|+C