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

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

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

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

\frac{1}{2} ln ∣ 2 ln (x) + 1 ∣ + C

Solution steps

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

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

= \int \frac{x - x ln ( x ^{2} )}{x ^{2} - ( x ln ( x ^{2} ) ) ^{2}} d x

Simplify

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

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

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = ln (x^{2}) + 1

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

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

= \frac{1}{2} ln ∣ 2 ln (x) + 1 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 ln (x) + 1 ∣ + C

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