What is the integral of 1/(xsqrt(64+ln^2(x))) ?

The integral of 1/(xsqrt(64+ln^2(x))) is ln(1/8 |ln(x)+sqrt(64+ln^2(x))|)+C

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integral of 1/(xsqrt(64+ln^2(x)))

\int \frac{1}{x 64 + ln ^{2} ( x )} d x

ln (\frac{1}{8} ln (x) + 64 + ln^{2} (x)) + C

Solution steps

\int \frac{1}{x 64 + ln ^{2} ( x )} d x

Apply u-substitution

: \int \frac{1}{64 + u ^{2}} d u

= \int \frac{1}{64 + u ^{2}} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= ln ∣ tan (v) + sec (v) ∣

Substitute back

= ln tan (arctan (\frac{1}{8} ln (x))) + sec (arctan (\frac{1}{8} ln (x)))

Simplify

ln tan (arctan (\frac{1}{8} ln (x))) + sec (arctan (\frac{1}{8} ln (x))) : ln (\frac{1}{8} ln (x) + 64 + ln^{2} (x))

= ln (\frac{1}{8} ln (x) + 64 + ln^{2} (x))

Add a constant to the solution

= ln (\frac{1}{8} ln (x) + 64 + ln^{2} (x)) + C

\int \frac{( 2 x - 3 )}{9 - x ^{2}} d x

x \to - \infty lim (e^{2} x^{2})

\frac{d}{d x} (e^{- 2 x} (cos (3 x) + sin (3 x)))

\int \frac{4400 q}{4 q ^{2} + 320} d q

y^{^{'}} = - 0.5 y^{1.5}

What is the integral of 1/(xsqrt(64+ln^2(x))) ?
The integral of 1/(xsqrt(64+ln^2(x))) is ln(1/8 |ln(x)+sqrt(64+ln^2(x))|)+C