What is the integral of (5ln(x))/(xsqrt(4-ln^4(x))) ?

The integral of (5ln(x))/(xsqrt(4-ln^4(x))) is 5/2 arcsin(1/2 ln^2(x))+C

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integral of (5ln(x))/(xsqrt(4-ln^4(x)))

\int \frac{5 ln ( x )}{x 4 - ln ^{4} ( x )} d x

\frac{5}{2} arcsin (\frac{1}{2} ln^{2} (x)) + C

Solution steps

\int \frac{5 ln ( x )}{x 4 - ln ^{4} ( x )} d x

Take the constant out:

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

= 5 \cdot \int \frac{ln ( x )}{x 4 - ln ^{4} ( x )} d x

Apply u-substitution

: \int \frac{u}{4 - u ^{4}} d u

= 5 \cdot \int \frac{u}{4 - u ^{4}} d u

Apply u-substitution

: \int \frac{1}{2 4 - v ^{2}} d v

= 5 \cdot \int \frac{1}{2 4 - v ^{2}} d v

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{4 - v ^{2}} d v

Apply Trigonometric Substitution

: \int 1 d w

= 5 \cdot \frac{1}{2} \cdot \int 1 d w

Integral of a constant:

\int a d x = a x

= 5 \cdot \frac{1}{2} \cdot 1 \cdot w

Substitute back

= 5 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{1}{2} ln^{2} (x))

Simplify

5 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{1}{2} ln^{2} (x)) : \frac{5}{2} arcsin (\frac{1}{2} ln^{2} (x))

= \frac{5}{2} arcsin (\frac{1}{2} ln^{2} (x))

Add a constant to the solution

= \frac{5}{2} arcsin (\frac{1}{2} ln^{2} (x)) + C

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What is the integral of (5ln(x))/(xsqrt(4-ln^4(x))) ?
The integral of (5ln(x))/(xsqrt(4-ln^4(x))) is 5/2 arcsin(1/2 ln^2(x))+C