What is the integral of 1/(x(7+ln(x^2))) ?

The integral of 1/(x(7+ln(x^2))) is 1/2 ln|7+2ln(x)|+C

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integral of 1/(x(7+ln(x^2)))

\int \frac{1}{x ( 7 + ln ( x ^{2} ) )} d x

\frac{1}{2} ln ∣ 7 + 2 ln (x) ∣ + C

Solution steps

\int \frac{1}{x ( 7 + ln ( x ^{2} ) )} d x

Apply u-substitution

: \int \frac{1}{2 u ( 7 + ln ( u ) )} d u

= \int \frac{1}{2 u ( 7 + ln ( 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 ( 7 + ln ( u ) )} d u

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

= \frac{1}{2} ln 7 + ln (x^{2})

Apply log rule

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

assuming

x \geq 0

= \frac{1}{2} ln ∣ 7 + 2 ln (x) ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 7 + 2 ln (x) ∣ + C

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What is the integral of 1/(x(7+ln(x^2))) ?
The integral of 1/(x(7+ln(x^2))) is 1/2 ln|7+2ln(x)|+C