What is the integral from 0 to 3/4 of 6/(x(ln(x))^3) ?

The integral from 0 to 3/4 of 6/(x(ln(x))^3) is -3/(ln^2(3/4))

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integral from 0 to 3/4 of 6/(x(ln(x))^3)

\int_{0}^{\frac{3}{4}} \frac{6}{x ( ln ( x ) ) ^{3}} d x

- \frac{3}{ln ^{2} ( \frac{3}{4} )}

Decimal Notation

- 36.24896 \dots

Solution steps

\int_{0}^{\frac{3}{4}} \frac{6}{x ( ln ( x ) ) ^{3}} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{\frac{3}{4}} \frac{1}{x ( ln ( x ) ) ^{3}} d x

Apply Integration By Parts

: \frac{1 - ln ^{3} ( x ) \cdot \int - \frac{l n ( x ) + 3}{x l n ^{4} ( x )} d x}{ln ^{3} ( x )}_{0}^{\frac{3}{4}}

= 6 \frac{1 - ln ^{3} ( x ) \cdot \int - \frac{l n ( x ) + 3}{x l n ^{4} ( x )} d x}{ln ^{3} ( x )}_{0}^{\frac{3}{4}}

\int - \frac{ln ( x ) + 3}{x ln ^{4} ( x )} d x = \frac{1}{2 ln ^{2} ( x )} + \frac{1}{ln ^{3} ( x )}

= 6 \frac{1 - ln ^{3} ( x ) ( \frac{1}{2 l n ^{2} ( x )} + \frac{1}{l n ^{3} ( x )} )}{ln ^{3} ( x )}_{0}^{\frac{3}{4}}

\frac{1 - ln ^{3} ( x ) ( \frac{1}{2 l n ^{2} ( x )} + \frac{1}{l n ^{3} ( x )} )}{ln ^{3} ( x )} = - \frac{1}{2 ln ^{2} ( x )}

= 6 [- \frac{1}{2 ln ^{2} ( x )}]_{0}^{\frac{3}{4}}

Compute the boundaries:

- \frac{1}{2 ln ^{2} ( \frac{3}{4} )}

= 6 (- \frac{1}{2 ln ^{2} ( \frac{3}{4} )})

Simplify

= - \frac{3}{ln ^{2} ( \frac{3}{4} )}

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What is the integral from 0 to 3/4 of 6/(x(ln(x))^3) ?
The integral from 0 to 3/4 of 6/(x(ln(x))^3) is -3/(ln^2(3/4))