What is the integral of (ln(x^{7/2}))/x ?

The integral of (ln(x^{7/2}))/x is 7/2 ln(x)ln|x|-7/4 ln|x|^2+C

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integral of (ln(x^{7/2}))/x

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

\frac{7}{2} ln (x) ln ∣ x ∣ - \frac{7}{4} ln ∣ x ∣^{2} + C

Solution steps

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

Apply Integration By Parts

: ln (x^{\frac{7}{2}}) ln ∣ x ∣ - \int \frac{7 ln ∣ x ∣}{2 x} d x

= ln (x^{\frac{7}{2}}) ln ∣ x ∣ - \int \frac{7 ln ∣ x ∣}{2 x} d x

\int \frac{7 ln ∣ x ∣}{2 x} d x = \frac{7}{4} ln ∣ x ∣^{2}

= ln (x^{\frac{7}{2}}) ln ∣ x ∣ - \frac{7}{4} ln ∣ x ∣^{2}

Apply log rule

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

assuming

x \geq 0

= \frac{7}{2} ln (x) ln ∣ x ∣ - \frac{7}{4} ln ∣ x ∣^{2}

Add a constant to the solution

= \frac{7}{2} ln (x) ln ∣ x ∣ - \frac{7}{4} ln ∣ x ∣^{2} + C

\frac{d}{d x} ((ln (x))^{p (x)})

\frac{d y}{d x} = \frac{y}{( x ^{2} + 1 )}

\frac{16}{s ( s + 2 ) ^{4}}

x \to \infty lim (\frac{e ^{- s x}}{s})

\frac{d y}{d x} y = (8 x^{2} - 16 + 16) e^{x}

What is the integral of (ln(x^{7/2}))/x ?
The integral of (ln(x^{7/2}))/x is 7/2 ln(x)ln|x|-7/4 ln|x|^2+C