What is the integral of 1/(3x^2sqrt(25-x^2)) ?

The integral of 1/(3x^2sqrt(25-x^2)) is -(sqrt(25-x^2))/(75x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(3x^2sqrt(25-x^2))

\int \frac{1}{3 x ^{2} 25 - x ^{2}} d x

- \frac{25 - x ^{2}}{75 x} + C

Solution steps

\int \frac{1}{3 x ^{2} 25 - x ^{2}} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{x ^{2} 25 - x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{25 cos ( u ) sin ( u )} d u

= \frac{1}{3} \cdot \int \frac{cot ( u )}{25 cos ( u ) sin ( u )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{25} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

\frac{cot ( u )}{cos ( u ) sin ( u )} : csc^{2} (u)

= \frac{1}{3} \cdot \frac{1}{25} \cdot \int csc^{2} (u) d u

Use the common integral:

\int csc^{2} (u) d u = - cot (u)

= \frac{1}{3} \cdot \frac{1}{25} (- cot (u))

Substitute back

u = arcsin (\frac{1}{5} x)

= \frac{1}{3} \cdot \frac{1}{25} (- cot (arcsin (\frac{1}{5} x)))

Simplify

\frac{1}{3} \cdot \frac{1}{25} (- cot (arcsin (\frac{1}{5} x))) : - \frac{25 - x ^{2}}{75 x}

= - \frac{25 - x ^{2}}{75 x}

Add a constant to the solution

= - \frac{25 - x ^{2}}{75 x} + C

\int \frac{x}{5 - x} d x

\int (tan (θ) + sec (θ))^{2} d θ

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

\int 2 1 - 25 x^{2} d x

\int x ln (4 + x^{2}) d x

What is the integral of 1/(3x^2sqrt(25-x^2)) ?
The integral of 1/(3x^2sqrt(25-x^2)) is -(sqrt(25-x^2))/(75x)+C