What is the integral of (2sqrt(ln^2(x)+16))/x ?

The integral of (2sqrt(ln^2(x)+16))/x is ln(x)sqrt(16+ln^2(x))+16ln(1/4 |ln(x)+sqrt(16+ln^2(x))|)+C

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integral of (2sqrt(ln^2(x)+16))/x

\int \frac{2 ln ^{2} ( x ) + 16}{x} d x

ln (x) 16 + ln^{2} (x) + 16 ln (\frac{1}{4} ln (x) + 16 + ln^{2} (x)) + C

Solution steps

\int \frac{2 ln ^{2} ( x ) + 16}{x} d x

Take the constant out:

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

= 2 \cdot \int \frac{ln ^{2} ( x ) + 16}{x} d x

Apply u-substitution

= 2 \cdot \int u^{2} + 16 d u

Apply Trigonometric Substitution

= 2 \cdot \int 16 sec^{3} (v) d v

Take the constant out:

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

= 2 \cdot 16 \cdot \int sec^{3} (v) d v

Apply Integral Reduction

= 2 \cdot 16 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} \cdot \int sec (v) d v)

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

= 2 \cdot 16 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= 2 \cdot 16 (\frac{sec ^{2} ( arctan ( \frac{1}{4} ln ( x ) ) ) sin ( arctan ( \frac{1}{4} ln ( x ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{4} ln (x))) + sec (arctan (\frac{1}{4} ln (x))))

Simplify

2 \cdot 16 (\frac{sec ^{2} ( arctan ( \frac{1}{4} ln ( x ) ) ) sin ( arctan ( \frac{1}{4} ln ( x ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{4} ln (x))) + sec (arctan (\frac{1}{4} ln (x)))) : ln (x) 16 + ln^{2} (x) + 16 ln (\frac{1}{4} ln (x) + 16 + ln^{2} (x))

= ln (x) 16 + ln^{2} (x) + 16 ln (\frac{1}{4} ln (x) + 16 + ln^{2} (x))

Add a constant to the solution

= ln (x) 16 + ln^{2} (x) + 16 ln (\frac{1}{4} ln (x) + 16 + ln^{2} (x)) + C

What is the integral of (2sqrt(ln^2(x)+16))/x ?
The integral of (2sqrt(ln^2(x)+16))/x is ln(x)sqrt(16+ln^2(x))+16ln(1/4 |ln(x)+sqrt(16+ln^2(x))|)+C