What is the integral of k/(x(ln(x))^b) ?

The integral of k/(x(ln(x))^b) is k/(-b+1)ln^{-b+1}(x)+C

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integral of k/(x(ln(x))^b)

\int \frac{k}{x ( ln ( x ) ) ^{b}} d x

\frac{k}{- b + 1} ln^{- b + 1} (x) + C

Solution steps

\int \frac{k}{x ( ln ( x ) ) ^{b}} d x

Take the constant out:

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

= k \cdot \int \frac{1}{x ( ln ( x ) ) ^{b}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

= k \cdot \int (ln (x))^{- b} \frac{1}{x} d x

Simplify

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

= k \cdot \int \frac{ln ^{- b} ( x )}{x} d x

Apply u-substitution

: \int u^{- b} d u

= k \cdot \int u^{- b} d u

Apply the Power Rule

: \frac{u ^{- b + 1}}{- b + 1}

= k \frac{u ^{- b + 1}}{- b + 1}

Substitute back

u = ln (x)

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

Simplify

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

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

Add a constant to the solution

= \frac{k}{- b + 1} ln^{- b + 1} (x) + C

x \to 0 + lim (\frac{e ^{4 x}}{3 x})

x \to 1 - lim (\frac{x ^{3}}{x ^{2} - 1})

(0, 0), (9, \frac{9}{7})

\int \frac{x ^{3} + x + 1}{x ^{4} + 2 x ^{2} + 1} d x

160 + 3.2 x - 0.03 x^{2}

What is the integral of k/(x(ln(x))^b) ?
The integral of k/(x(ln(x))^b) is k/(-b+1)ln^{-b+1}(x)+C