What is the integral of 1/(x^2+5[2x^2-3]) ?

The integral of 1/(x^2+5[2x^2-3]) is -1/(2sqrt(165))(ln|sqrt(11/15)x+1|-ln|sqrt(11/15)x-1|)+C

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integral of 1/(x^2+5[2x^2-3])

integral

\frac{1}{x ^{2} + 5 [ 2 x ^{2} - 3 ]}

- \frac{1}{2 165} (ln \frac{11}{15} x + 1 - ln \frac{11}{15} x - 1) + C

Solution steps

\int \frac{1}{x ^{2} + 5 [ 2 x ^{2} - 3 ]} d x

Expand

x^{2} + 5 (2 x^{2} - 3) : 11 x^{2} - 15

= \int \frac{1}{11 x ^{2} - 15} d x

Apply Integral Substitution

= \int \frac{1}{165 ( u ^{2} - 1 )} d u

Take the constant out:

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

= \frac{1}{165} \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

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

= \frac{1}{165} \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

= \frac{1}{165} (- \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}

= \frac{1}{165} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{11}{15} x

= \frac{1}{165} - \frac{ln \frac{11}{15} x + 1}{2} - \frac{ln \frac{11}{15} x - 1}{2}

Simplify

\frac{1}{165} - \frac{ln \frac{11}{15} x + 1}{2} - \frac{ln \frac{11}{15} x - 1}{2} : - \frac{1}{2 165} (ln \frac{11}{15} x + 1 - ln \frac{11}{15} x - 1)

= - \frac{1}{2 165} (ln \frac{11}{15} x + 1 - ln \frac{11}{15} x - 1)

Add a constant to the solution

= - \frac{1}{2 165} (ln \frac{11}{15} x + 1 - ln \frac{11}{15} x - 1) + C

\frac{5 x \frac{2}{3} + 2}{3 x ^{2} + 5 x}

4 x^{2} - 62 x

(x^{2} + 1)^{0.5}

\frac{1}{( x + a )}

(ln (x))^{3}

What is the integral of 1/(x^2+5[2x^2-3]) ?
The integral of 1/(x^2+5[2x^2-3]) is -1/(2sqrt(165))(ln|sqrt(11/15)x+1|-ln|sqrt(11/15)x-1|)+C