What is the integral of (16)/(xsqrt(1+2x)) ?

The integral of (16)/(xsqrt(1+2x)) is -32(1/2 ln|sqrt(1+2x)+1|-1/2 ln|sqrt(1+2x)-1|)+C

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integral of (16)/(xsqrt(1+2x))

\int \frac{16}{x 1 + 2 x} d x

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

Solution steps

\int \frac{16}{x 1 + 2 x} d x

Take the constant out:

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

= 16 \cdot \int \frac{1}{x 1 + 2 x} d x

Apply u-substitution

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

= 16 \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

= 16 \cdot 2 \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

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

= 16 \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

= 16 \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}

= 16 \cdot 2 (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = 1 + 2 x

= 16 \cdot 2 (- (\frac{ln 1 + 2 x + 1}{2} - \frac{ln 1 + 2 x - 1}{2}))

Simplify

16 \cdot 2 (- (\frac{ln 1 + 2 x + 1}{2} - \frac{ln 1 + 2 x - 1}{2})) : - 32 (\frac{1}{2} ln 1 + 2 x + 1 - \frac{1}{2} ln 1 + 2 x - 1)

= - 32 (\frac{1}{2} ln 1 + 2 x + 1 - \frac{1}{2} ln 1 + 2 x - 1)

Add a constant to the solution

= - 32 (\frac{1}{2} ln 1 + 2 x + 1 - \frac{1}{2} ln 1 + 2 x - 1) + C

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What is the integral of (16)/(xsqrt(1+2x)) ?
The integral of (16)/(xsqrt(1+2x)) is -32(1/2 ln|sqrt(1+2x)+1|-1/2 ln|sqrt(1+2x)-1|)+C