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

The integral of 2/(xsqrt(1+2x)) is -4(1/2 ln|sqrt(1+2x)+1|-1/2 ln|sqrt(1+2x)-1|)+C

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Factor

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

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

Substitute back

u = 1 + 2 x

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{t ^{2} - 2 t + 37} d t

\int_{0}^{\frac{π}{4}} 7 sec^{2} (x) d x

\frac{d}{d x} (3 (2 x e^{x} + e^{x} x^{2}))

y^{^{'}} = \frac{y ^{2}}{t ^{3}}

ln (x e^{- 6 x})

What is the integral of 2/(xsqrt(1+2x)) ?
The integral of 2/(xsqrt(1+2x)) is -4(1/2 ln|sqrt(1+2x)+1|-1/2 ln|sqrt(1+2x)-1|)+C