What is the integral of 4/(3x^2sqrt(5-7x^2)) ?

The integral of 4/(3x^2sqrt(5-7x^2)) is -(4sqrt(5-7x^2))/(15x)+C

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integral of 4/(3x^2sqrt(5-7x^2))

\int \frac{4}{3 x ^{2} 5 - 7 x ^{2}} d x

- \frac{4 5 - 7 x ^{2}}{15 x} + C

Solution steps

\int \frac{4}{3 x ^{2} 5 - 7 x ^{2}} d x

Take the constant out:

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

= \frac{4}{3} \cdot \int \frac{1}{x ^{2} 5 - 7 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{7}{5 sin ^{2} ( u )} d u

= \frac{4}{3} \cdot \int \frac{7}{5 sin ^{2} ( u )} d u

Take the constant out:

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

= \frac{4}{3} \cdot \frac{7}{5} \cdot \int \frac{1}{sin ^{2} ( u )} d u

Use the common integral:

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

= \frac{4}{3} \cdot \frac{7}{5} (- cot (u))

Substitute back

u = arcsin (\frac{7}{5} x)

= \frac{4}{3} \cdot \frac{7}{5} (- cot (arcsin (\frac{7}{5} x)))

Simplify

\frac{4}{3} \cdot \frac{7}{5} (- cot (arcsin (\frac{7}{5} x))) : - \frac{4 5 - 7 x ^{2}}{15 x}

= - \frac{4 5 - 7 x ^{2}}{15 x}

Add a constant to the solution

= - \frac{4 5 - 7 x ^{2}}{15 x} + C

\frac{d}{d x} (\frac{4 x}{3 - cot ( x )})

f (x) = 6 ln (x)

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

\frac{d}{d x} ((\frac{1}{3 x + 1})^{4})

What is the integral of 4/(3x^2sqrt(5-7x^2)) ?
The integral of 4/(3x^2sqrt(5-7x^2)) is -(4sqrt(5-7x^2))/(15x)+C