Find the indefinite integral ∫ln⁡(x3)dx (a) using a table of integrals and (b) using the Integration by parts method.

Given, ∫ln⁡(x3)dx a). By using the table of integrals, we have ∫ln⁡xdx=xln⁡(x)−x+C Therefore, ∫ln⁡(x3)dx=x3×ln⁡(x3)−x313+C (x-coefficient must be divided) =xln⁡(x3)−x+C Step 2 b). ∫ln⁡(x3)dx=∫ln⁡(x3)×1dx Now integrating by parts, we get ∫ln⁡(x3)×1dx=ln⁡(x3)×∫1dx−∫[ddx[ln⁡(x3)]×∫1dx]+C =ln⁡(x3)×x−∫(1x3×13×x)dx+C =ln⁡(x3)×x−∫1dx+C =xln⁡(x3)−x+C

Expert answers to "Find the indefinite integral ∫ln⁡(x3)dx (a) using a..."

rocedwrp

Answered question

2021-05-23

Find the indefinite integral

\int \ln (\frac{x}{3}) d x

(a) using a table of integrals and (b) using the Integration by parts method.

Answer & Explanation

tafzijdeq

Skilled2021-05-24Added 92 answers

Given,
$\int \ln (\frac{x}{3}) d x$
a).
By using the table of integrals, we have $\int \ln x d x = x \ln (x) - x + C$
Therefore,
$\int \ln (\frac{x}{3}) d x = \frac{\frac{x}{3} \times \ln (\frac{x}{3}) - \frac{x}{3}}{\frac{1}{3}} + C$ (x-coefficient must be divided)
$= x \ln (\frac{x}{3}) - x + C$
Step 2
b).
$\int \ln (\frac{x}{3}) d x = \int \ln (\frac{x}{3}) \times 1 d x$
Now integrating by parts, we get
$\int \ln (\frac{x}{3}) \times 1 d x = \ln (\frac{x}{3}) \times \int 1 d x - \int [\frac{d}{d x} [\ln (\frac{x}{3})] \times \int 1 d x] + C$
$= \ln (\frac{x}{3}) \times x - \int (\frac{1}{\frac{x}{3}} \times \frac{1}{3} \times x) d x + C$
$= \ln (\frac{x}{3}) \times x - \int 1 d x + C$
$= x \ln (\frac{x}{3}) - x + C$

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