Question

Integral and simplifying logarithms

Answer 1

i need a min to type the equation

Answer 2

k

Answer 3

\[\int\limits_{0}^{1} (x+1) + \frac{ 1 }{ x-3 }+\frac{ 5 }{ x+2 }\]

Answer 4

= \[\frac{ 1 }{ 2 } x^2 + x + \ln \left| x -3\right|+ 5\ln \left| x+2\right| \]

Answer 5

from zero to one

Answer 6

im sorrry i have no idea

Answer 7

= \[(\frac{ 3 }{ 2 } +\ln \left| -2 \right|+5 \ln \left| 3 \right|)-(\frac{ 3 }{ 2 } + \ln \left| -3 \right|+5\ln \left| 2 \right|)\]

Answer 8

this is where I get stuck. Can anyone help me simplify?

Answer 9

i have no clue im srry

Answer 10

HI

Answer 11

what do you need, a number?

Answer 12

it is a calculator exercise now, you will not compute this with pencil and paper

Answer 13

your integral is ok:
\[ \frac{ 1 }{ 2 } x^2 + x + \ln \left| x -3\right|+ 5\ln \left| x+2\right| \]
at the upper limit, x=1 you get
\[ \frac{ 1 }{ 2 } + 1 + \ln | 1-3| + 5 \ln | 1+2 | \\
= \frac{3}{2} + \ln 2 + 5 \ln 3
\]
at the lower limit x=0 you get
\[ 0 + 0 + \ln | -3| + 5 \ln|2| \\
= \ln 3 + 5 \ln 2
\]
subtracting upper - lower:
\[ \frac{3}{2} + \ln 2 + 5 \ln 3 - \ln 3 - 5 \ln 2 \\
= \frac{3}{2} + 4\ln 3 - 4 \ln 2
\]
you can combine / re-write the logs using
\[ \ln(a^b) = b \ln(a) \\ \ln \left(\frac{a}{b}\right) = \ln(a) - \ln(b) \]
in other words:
\[ \frac{3}{2} + 4\ln 3 - 4 \ln 2 \\
=\frac{3}{2} + \ln 3^4 - \ln 2^4
=\frac{3}{2} + \ln 81 - \ln 16\\
= \frac{3}{2} +\ln\frac{81}{16}
\]

Answer 14

it says the correct final is 3/2 - 4/5 ln 3/2

Answer 15

\[\frac{ 3 }{ 2 }-\frac{ 4 }{ 5 }\ln \frac{ 2 }{ 3 }\]

Answer 16

i got what you did Phi...it says i am wrong and showed me that answeer

Answer 17

ok, we can do this
\[ \frac{3}{2} + 4 \ln \frac{3}{2} \\= \frac{3}{2} - 4 \ln \frac{2}{3} \]
but I don't see where the 1/5 comes from

Answer 18

I would ask your teacher about this, because the answer does not match up with the question.