Question

Evaluate the following integral (Calculus 2) Please show step by step explanation!

Answer 1

Answer:

$\displaystyle \int \dfrac{\ln x}{x^8}\:\:\text{d}x=-\dfrac{\ln x}{7x^7}- \dfrac{1}{49x^7}+\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given integral:

$\displaystyle \int \dfrac{\ln x}{x^8}\:\:\text{d}x$

$\textsf{Apply exponent rule} \quad \dfrac{1}{a^n}=a^{-n}$

$\implies \displaystyle \int x^{-8}\ln x\:\:\text{d}x$

$\boxed{\begin{minipage}{4.8 cm}\underline{Integrating x^n }\\\\ \displaystyle \int x^n\:\text{d}x=\dfrac{x^{n+1}}{n+1}+\text{C} \end{minipage}}$

$\boxed{\begin{minipage}{4.8 cm}\underline{Integration by parts} \\\\ \displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x \\ \end{minipage}}$

Using Integration by parts:

$\textsf{Let }u=\ln x \implies \dfrac{\text{d}u}{\text{d}x}=\dfrac{1}{x}$

$\textsf{Let }\dfrac{\text{d}v}{\text{d}x}=x^{-8} \implies v=-\dfrac{1}{7}x^{-8+1}=-\dfrac{1}{7}x^{-7}=-\dfrac{1}{7x^7}$

Therefore:

$\begin{aligned}\displaystyle \int u \dfrac{dv}{dx}\:dx & =uv-\int v\: \dfrac{du}{dx}\:dx\\\\\implies \displaystyle \int x^{-8}\ln x\:\:\text{d}x & = -\dfrac{1}{7x^7}\ln x- \int -\dfrac{1}{7x^7} \cdot \dfrac{1}{x}\:\:\text{d}x\\\\& = -\dfrac{\ln x}{7x^7}+ \int \dfrac{1}{7x^8}\:\:\text{d}x\\\\& = -\dfrac{\ln x}{7x^7}+ \int \dfrac{1}{7}x^{-8}\:\:\text{d}x\\\\& = -\dfrac{\ln x}{7x^7}-\dfrac{1}{7 \cdot 7}x^{-8+1}+\text{C}\\\\& = -\dfrac{\ln x}{7x^7}- \dfrac{1}{49x^7}+\text{C}\end{aligned}$

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Answer 2

Substitute $y=\ln(x) \iff e^y = x$ and $dy=\frac{dx}x$ to get

$\displaystyle \int \frac{\ln(x)}{x^8} \, dx = \int ye^{-7y} \, dy$

Integrate by parts with

$u = y \implies du = dy$

$dv = e^{-7y} \, dy \implies v = -\dfrac17 e^{-7y}$

Then

$\displaystyle \int u\,dv = uv - \int v\,du \\\\ \implies \int y e^{-7y} \, dy = -\dfrac17 ye^{-7y} + \frac17 \int e^{-7y} \, dy + C \\\\ ~~~~~~~~~~~~ = -\frac17 ye^{-7y} - \frac1{49} e^{-7y} + C \\\\ ~~~~~~~~~~~~ = \boxed{-\frac{\ln(x)}{7x^7} - \frac1{49x^7} + C}$