What is the integral of 4/(xsqrt(16-ln(x^2))) ?

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

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integral of 4/(xsqrt(16-ln(x^2)))

\int \frac{4}{x 16 - ln ( x ^{2} )} d x

- 4 16 - 2 ln (x) + C

Solution steps

\int \frac{4}{x 16 - ln ( x ^{2} )} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{x 16 - ln ( x ^{2} )} d x

Apply u-substitution

: \int \frac{1}{2 u 16 - ln ( u )} d u

= 4 \cdot \int \frac{1}{2 u 16 - ln ( u )} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{u 16 - ln ( u )} d u

Apply u-substitution

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

= 4 \cdot \frac{1}{2} \cdot \int - \frac{1}{v} d v

Take the constant out:

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

= 4 \cdot \frac{1}{2} (- \int \frac{1}{v} d v)

Apply the Power Rule

: 2 v^{\frac{1}{2}}

= 4 \cdot \frac{1}{2} (- 2 v^{\frac{1}{2}})

Substitute back

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

Simplify

4 \cdot \frac{1}{2} (- 2 (16 - ln (x^{2}))^{\frac{1}{2}}) : - 4 16 - 2 ln (x)

= - 4 16 - 2 ln (x)

Add a constant to the solution

= - 4 16 - 2 ln (x) + C

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