What is the integral of 1/(xsqrt(4-9ln^2(x))) ?

The integral of 1/(xsqrt(4-9ln^2(x))) is 1/3 arcsin(3/2 ln(x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(xsqrt(4-9ln^2(x)))

\int \frac{1}{x 4 - 9 ln ^{2} ( x )} d x

\frac{1}{3} arcsin (\frac{3}{2} ln (x)) + C

Solution steps

\int \frac{1}{x 4 - 9 ln ^{2} ( x )} d x

Apply u-substitution

: \int \frac{1}{4 - 9 u ^{2}} d u

= \int \frac{1}{4 - 9 u ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{1}{3} d v

= \int \frac{1}{3} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot v

Substitute back

= \frac{1}{3} \cdot 1 \cdot arcsin (\frac{3}{2} ln (x))

Simplify

= \frac{1}{3} arcsin (\frac{3}{2} ln (x))

Add a constant to the solution

= \frac{1}{3} arcsin (\frac{3}{2} ln (x)) + C

f (x) = sin (2 x), at x = - \frac{π}{3}

\frac{( x ^{2} - 1 ) ^{3}}{x ^{2} + 2}

\int (- sec^{2} (x)) d x

h (x) = (ln (x)) (3 x^{3})

\int e^{4 x + 1} d x

What is the integral of 1/(xsqrt(4-9ln^2(x))) ?
The integral of 1/(xsqrt(4-9ln^2(x))) is 1/3 arcsin(3/2 ln(x))+C