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

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

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integral of 1/(xsqrt(1+2x))

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

- ln 1 + 2 x + 1 + ln 1 + 2 x - 1 + C

Solution steps

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

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

Simplify

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

= - ln 1 + 2 x + 1 + ln 1 + 2 x - 1

Add a constant to the solution

= - ln 1 + 2 x + 1 + ln 1 + 2 x - 1 + C

y^{^{'}} - y = e^{- t} sin (t)

n = 1 \sum \infty ln (n^{2} + 2 n)

n = 1 \sum \infty \frac{1}{7 ^{n}}

\int (x - 2)^{2} sin (x) d x

\int \frac{1}{x ^{2} + 3 x - 4} d x

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