What is the integral of 5/(3x^2-2) ?

The integral of 5/(3x^2-2) is -5/(2sqrt(6))(ln|sqrt(3/2)x+1|-ln|sqrt(3/2)x-1|)+C

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integral of 5/(3x^2-2)

\int \frac{5}{3 x ^{2} - 2} d x

- \frac{5}{2 6} (ln \frac{3}{2} x + 1 - ln \frac{3}{2} x - 1) + C

Solution steps

\int \frac{5}{3 x ^{2} - 2} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{3 x ^{2} - 2} d x

Factor

3 x^{2} - 2 : - (- 3 x^{2} + 2)

= 5 \cdot \int \frac{1}{- ( - 3 x ^{2} + 2 )} d x

Take the constant out:

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

= 5 (- \int \frac{1}{- 3 x ^{2} + 2} d x)

Apply Integral Substitution

= 5 (- \int \frac{1}{6 ( - u ^{2} + 1 )} d u)

Take the constant out:

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

= 5 (- \frac{1}{6} \cdot \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}

= 5 (- \frac{1}{6} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{3}{2} x

= 5 - \frac{1}{6} \frac{ln \frac{3}{2} x + 1}{2} - \frac{ln \frac{3}{2} x - 1}{2}

Simplify

5 - \frac{1}{6} \frac{ln \frac{3}{2} x + 1}{2} - \frac{ln \frac{3}{2} x - 1}{2} : - \frac{5}{2 6} (ln \frac{3}{2} x + 1 - ln \frac{3}{2} x - 1)

= - \frac{5}{2 6} (ln \frac{3}{2} x + 1 - ln \frac{3}{2} x - 1)

Add a constant to the solution

= - \frac{5}{2 6} (ln \frac{3}{2} x + 1 - ln \frac{3}{2} x - 1) + C

x \to \infty lim (- 5^{x})

- 3

- \frac{2 sin ( 4 x )}{x}

y = 8 cos (6 x), y = 8 - 8 cos (6 x)

cos (8 x)

What is the integral of 5/(3x^2-2) ?
The integral of 5/(3x^2-2) is -5/(2sqrt(6))(ln|sqrt(3/2)x+1|-ln|sqrt(3/2)x-1|)+C