Question

CALCULUS - NATURAL LOGS AND INTEGRATION WITH DEFINITE INTERVALS

Answer 1

\[\int\limits_{1}^{e} \frac{ (1+\ln x)^2 }{ x }\]

Answer 2

evaluate the definite integral. I tried taking a stab at it by substitution by letting u= 1+lnx and du=1/x but i got stuck there

Answer 3

any ideas?

Answer 4

Gimme a moment

Answer 5

sure

Answer 6

forgot to add dx btw

Answer 7

Let's try u-sub
\[\int\limits\limits_{1}^{e} \frac{ (1+\ln x)^2 }{ x }dx\]
Let \(u=ln(x)\), \(du=\frac{1}{x}dx\)
This makes the lower bound \(u=ln(x)=ln(1)=0, \) and upper bound \(u=ln(e)=1.\)
\[\int\limits\limits_{0}^{1} (u+1)^2 du\]
u-sub again and it's done

Answer 8

um im lost. so then i would do \[[(\ln (1)+1)^2 - (\ln (0)+1)^2]\]

Answer 9

Hold on lemme write it myself

Answer 10

integrating the intervals 0 and 1 into f(a)-f(b) right? but my book shows \[\frac{ 7 }{ 3 }\] as the answer

Answer 11

alright

Answer 12

http://prnt.sc/dhson1

Answer 13

thanks!

Answer 14

Cheers

Answer 15

There is a pattern here, there often is, that obviates the need for any substitution

Note that

\(\dfrac{d}{dx} (1 + \ln x )^3 = 3 (1 + \ln x )^2 \cdot \dfrac{1}{x}\)


So you integral can be re-written as

\(\int\limits_1^e ~ \dfrac{d}{dx} (\frac{1}{3} (1 + \ln x )^3) ~ dx\)

\(= \left . \frac{1}{3} (1 + \ln x )^3 ~ \right|_1^e\)

That process also shows that the sub \(y = 1 + ln x, dy = \dfrac{1}{x} dx\) (so that \(dx = x ~ dy\) !!) is better ......... though clearly you don't need to plod through any substitution at all if you see the pattern :)