Question

int from 1 to 2 [xlnx]

Answer 1

\[\int\limits_{}^{}xln(x)dx=\frac{x^2}{2}\ln(x)-\int\limits_{}^{}\frac{x^2}{2} \cdot \frac{1}{x} dx+C\]
\[=\frac{x^2}{2}\ln(x)-\int\limits_{}^{}\frac{x}{2}dx=\frac{x^2}{2}\ln(x)-\frac{1}{2} \cdot \frac{x^2}{2}+C\]
\[=\frac{x^2}{2}(\ln(x)-\frac{1}{2})+C\]
Check:
\[[\frac{x^2}{2}(\ln(x)-\frac{1}{2})+C]'=x(\ln(x)-\frac{1}{2})+(\frac{1}{x}-0)\frac{x^2}{2}+0\]
\[=x \ln(x)-\frac{1}{2}x+\frac{x}{2}=x \ln(x)\]
now we have
\[\int\limits_{1}^{2}x \ln(x) dx=[\frac{x^2}{2}(\ln(x)-\frac{1}{2})]_1^{2}\]

Answer 2

\[=\frac{2^2}{2}(\ln(2)-\frac{1}{2})-\frac{1^2}{2}(\ln(1)-\frac{1}{2})=2(\ln(2)-\frac{1}{2})-\frac{1}{2}(0-\frac{1}{2})\]
\[=2 \ln(2)-1+\frac{1}{4}=2\ln(2)-\frac{4}{4}+\frac{1}{4}=2 \ln(2)-\frac{3}{4}\]

Answer 3

Cool! That's exactly what I got too..I was kind of tired of the repetitive algebra so I wanted to thrown some calculus in lol :)

Answer 4

throw*

Answer 5

try this one:
\[\int\limits_{}^{}e^{\sqrt{3s+9}}ds\]

Answer 6

okay

Answer 7

god my cat just shedded all over me

Answer 8

its time for a brushing

Answer 9

haha

Answer 10

This is what I got:\[\ \frac{2\sqrt{3s+9}}{3} \ e ^{\sqrt{3s+9}}+C\]

Answer 11

I see integration by parts was necessary in this case

Answer 12

\[u^2=3s+9=>2u du=3 ds\]
\[\frac{2}{3}\int\limits_{}^{}u e^u du=\frac{2}{3}(e^u u-\int\limits_{}^{}e^u du)+C=\frac{2}{3}e^u u-\frac{2}{3}e^u+C\]
\[=\frac{2}{3}e^{\sqrt{3s+9}}\sqrt{3s+9}-\frac{2}{3}e^{\sqrt{3s+9}}+C\]

Answer 13

\[=\frac{2}{3}e^{\sqrt{3s+9}}(\sqrt{3s+9}-1)+C\]

Answer 14

yeah I see it know

Answer 15

i think polpak showed me a way without integration by parts i can't remember

Answer 16

hmm..

Answer 17

Okay an

Answer 18

okay another one

Answer 19

another one?
ok let me think

Answer 20

\[\int\limits_{}^{}\ln(x+x^2)dx\]

Answer 21

ok gimme a few minutes i'm actually going to eat i'll be back

Answer 22

lol never mind I take that back I'll be back later

Answer 23

Alright sorry i took long: Here is my answer:\[x \ln(x+x ^{2}) + \ln(x+1)-2x+C\]