What is the integral of 1/(xsqrt(8-ln^2(x)+2ln(x))) ?

The integral of 1/(xsqrt(8-ln^2(x)+2ln(x))) is arcsin(1/3 (ln(x)-1))+C

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integral of 1/(xsqrt(8-ln^2(x)+2ln(x)))

\int \frac{1}{x 8 - ln ^{2} ( x ) + 2 ln ( x )} d x

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

Solution steps

\int \frac{1}{x 8 - ln ^{2} ( x ) + 2 ln ( x )} d x

Apply u-substitution

: \int \frac{1}{8 - u ^{2} + 2 u} d u

= \int \frac{1}{8 - u ^{2} + 2 u} d u

Complete the square

8 - u^{2} + 2 u : - (u - 1)^{2} + 9

= \int \frac{1}{- ( u - 1 ) ^{2} + 9} d u

Apply u-substitution

: \int \frac{1}{- v ^{2} + 9} d v

= \int \frac{1}{- v ^{2} + 9} d v

Apply Trigonometric Substitution

: \int 1 d w

= \int 1 d w

Integral of a constant:

\int a d x = a x

= 1 \cdot w

Substitute back

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

Simplify

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

Add a constant to the solution

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

x \to 0 + lim (6 x ln (x))

x \to - 2 lim (x^{2} - 4)

\frac{\partial}{\partial x} (2 (y - 5) (x + 5))

\frac{d}{d x} (arcsec (8 x))

\int - cos (x) + sin (x) + 5 d x

What is the integral of 1/(xsqrt(8-ln^2(x)+2ln(x))) ?
The integral of 1/(xsqrt(8-ln^2(x)+2ln(x))) is arcsin(1/3 (ln(x)-1))+C