What is the integral of (17)/(xsqrt(1+2x)) ?

The integral of (17)/(xsqrt(1+2x)) is -34(1/2 ln|sqrt(1+2x)+1|-1/2 ln|sqrt(1+2x)-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (17)/(xsqrt(1+2x))

\int \frac{17}{x 1 + 2 x} d x

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

Solution steps

\int \frac{17}{x 1 + 2 x} d x

Take the constant out:

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

= 17 \cdot \int \frac{1}{x 1 + 2 x} d x

Apply u-substitution

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

= 17 \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

= 17 \cdot 2 \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

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

= 17 \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

= 17 \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}

= 17 \cdot 2 (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = 1 + 2 x

= 17 \cdot 2 (- (\frac{ln 1 + 2 x + 1}{2} - \frac{ln 1 + 2 x - 1}{2}))

Simplify

17 \cdot 2 (- (\frac{ln 1 + 2 x + 1}{2} - \frac{ln 1 + 2 x - 1}{2})) : - 34 (\frac{1}{2} ln 1 + 2 x + 1 - \frac{1}{2} ln 1 + 2 x - 1)

= - 34 (\frac{1}{2} ln 1 + 2 x + 1 - \frac{1}{2} ln 1 + 2 x - 1)

Add a constant to the solution

= - 34 (\frac{1}{2} ln 1 + 2 x + 1 - \frac{1}{2} ln 1 + 2 x - 1) + C

\int \frac{x ^{2} + 2 x + 7}{x ^{2} + 6 x + 5} d x

\int \frac{5}{16 x ( ln ( x ) ) ^{8}} d x

\int \frac{3 x ^{2} - 3 x + 18}{x ^{3} + 3 x} d x

\int \frac{7 x - 23}{6 ( x - 2 ) ( x - 5 )} d x

\int e^{s i n (x) + 1} \cdot cos (x) d x

What is the integral of (17)/(xsqrt(1+2x)) ?
The integral of (17)/(xsqrt(1+2x)) is -34(1/2 ln|sqrt(1+2x)+1|-1/2 ln|sqrt(1+2x)-1|)+C