Question

Just doing me homework ....

Just doing me homework ....

@Mathematics

Answer 1

the answer to that question is homework is lame but we have to get it done lol

Answer 2

\[\Large
\begin{array}l
1)\ \int \frac{x}{x-6}\ dx

\end{array}
\]

Answer 3

I sooo agree

Answer 4

I dont know how to do that i dont even know what kind of math that is!

Answer 5

\[\Large
\begin{array}l
1)\ \int \frac{x}{x-6}\ dx\\\\
=\int \frac{x+6-6}{x-6}dx\\\\
=\int \frac{x-6}{x-6}+\frac{6}{x-6}dx\\\\
=\int 1dx+6\int \frac{1}{x-6}dx\\\\
=x+ 6ln|x-6|+C\\\\


\end{array}\]

Answer 6

calculus 2 ... boring stuff ;)

Answer 7

^^ how you format your picture for this site? I copied your image .. it's working but I want to know how to do it myself ?

Answer 8

\[\Large
\begin{array}l
9)\ \int \frac{x-9}{(x+5)(x-2)}\ dx\\\\
=\int \frac{1}{x+5}dx+B\int\frac{1}{x-2}dx\\\\
=A\ ln|x+5|+B\ ln|x-2|+C\\\\
\frac{x-9}{(x+5)(x-2)}=\frac{A}{x+5}+\frac{B}{x-2}\\\\
x-9=A(x-2)+B(x+5)\\\\
2-9=B(2+5);\ B=-1\\\\
-5-9=A(-5-2);\ A=2\\\\
=2\ ln|x+5|-\ ln|x-2|+C\\\\
\end{array}\]

Answer 9

the picture should be with a certain size; try to make it about the size that it would look and adjust it from there id say

Answer 10

my format is jpeg

Answer 11

and duplication is frowned upon as a rule of thumb. just a heads up ;)

Answer 12

change of colors or some such would be advisable ...

Answer 13

^^ but what tools did you used ?

Answer 14

windows paint. First I latexed me up the int xdx and hit the PrtScr button on the keyboard and pasted it into Paint

Answer 15

then dolled it up from there

Answer 16

I am on it .. I have my eyes on Euler-Mascheroni constant ;)

Answer 17

\Large
\begin{array}l
15)\ \int_{3}^{4} \frac{x^3-2x^2-4}{x^3-2x^2}\ dx\\\\
\end{array}

Answer 18

cool lol

Answer 19

\[\Large
\begin{array}l

15)\ \int_{3}^{4} \frac{x^3-2x^2-4}{x^3-2x^2}\ dx\\\\
=\int_{3}^{4} 1-\frac{4}{x^3-2x^2}\ dx\\\\
=\int_{3}^{4} 1-\frac{4}{x^2(x-2)}\ dx\\\\
\left.=x-(A\ ln(x)-B\ \frac{1}{x}+C\ ln(x-2))\ \right|_{3}^{4}\\\\
\frac{4}{x^2(x-2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-2}\\\\
4=A(x)(x-2)+B(x-2)+C(x^2)\\\\
4=C(2^2);\ C=1\\\\
4=B(0-2);\ B=-2\\\\
0=Ax^2+x^2;\ A=-1\\\\
\left.=x-(-\ ln(x)+\frac{2}{x}+\ ln(x-2))\ \right|_{3}^{4}\\\\
=4+ln(4)-\frac{2}{4}- ln(4-2)\\\\
-3-ln(3)+\frac{2}{3}+ ln(3-2)\\\\
=\frac{7}{6}+ln(\frac{2}{3})\ \approx.761202\\\\

\end{array}\]

Answer 20

\[\Large
\begin{array}l
20)\ \int \frac{x^2-5x+16}{(2x+1)(x-2)^2}\ dx\\\\
\frac{x^2-5x+16}{(2x+1)(x-2)^2}=\frac{A}{(2x+1)^1}\\\\+\frac{B}{(x-2)^1}+\frac{C}{(x-2)^2}\\\\
x^2-5x+16=A(x-2)^2+B(2x-2)(2x+1)\\\\+C(2x+1)\\\\
10=C(5);\ C=2\\\\
75=A(25);\ A=3\\\\
x^2=3x^2+B(2x^2);\ B=-1\\\\
\int 3\frac{1}{2x+1}-1\frac{1}{x-2}+2\frac{1}{(x-2)^2}\ dx\\\\
\frac{3}{2}ln|2x+1|-ln|x-2|-\frac{2}{x-2}+C\\\\
\end{array}\]

Answer 21

the preview lies; it doesnt take into account the size of the icon ..

Answer 22

\[\Large
\begin{array}l
22)\ \int \frac{1}{x^2(x-1)^2} dx\\\\
\frac{1}{x^2(x-1)^2}=\frac{c_1}{x}+\frac{c_2}{x^2}+\frac{c_3}{(x-1)}+\frac{c_4}{(x-1)^2}\\\\
1=c_1(x)(x-1)^2+c_2(x-1)^2\\\\+c_3(x^2)(x-1)+c_4x^2\\\\
1=c_4\\\\
1=c_2\\\\
0=c_1+c_3\\\\
-2=-c_1;\ c_1=2;\ c_3=-2\\\\
\int \frac{2}{x}+\frac{1}{x^2}+\frac{-2}{x-1}+\frac{1}{(x-1)^2}dx\\\\
2ln|x|-2ln|x-1|-\frac{1}{x}-\frac{1}{(x-1)}+C\\\\

\end{array}\]

Answer 23

\[\Large
\begin{array}l
34)\ \int \frac{x^3}{x^3+1} dx=\int \frac{x^3+1-1}{x^3+1} dx\\\\
=x-\int \frac{1}{(x+1)(x^2-x+1)} dx\\\\
\frac{1}{(x+1)(x^2-x+1)}=\frac{a_1}{x+1}+\frac{a_2x+a_3}{x^2-x+1}\\\\
1=a_1(x^2-x+1)+(a_2x+a_3)(x+1)\\\\
1=a_1x^2-a_1x+a_1+a_2x^2+a_2x+a_3x\\\\+a_3\\\\
1=x^2(a_1+a_2)+x(-a_1+a_2+a_3)\\\\+a_1+a_3\\\\
0=a_1+a_2;\ a_1=-a_2\\\\
0=2a_2+a_3\ a_3=-2a_2\\\\
1=-a_2-2a_2;\ a_2=-\frac{1}{3}\\\\
a_3=\frac{2}{3};\ a_1=\frac{1}{3}\\\\
=x-\int \frac{1}{3(x+1)}+\frac{-x+2}{3(x^2-x+1)}dx\\\\
=x-\int \frac{1}{3(x+1)}dx-\int\frac{-x+2}{3(x^2-x+1)}dx\\\\
=x-\frac{1}{3}ln|x+1|+\frac{1}{3}\int\frac{x-2}{x^2-x+1}dx\\\\
=x-\frac{1}{3}ln|x+1|+\frac{1}{6}\int\frac{2x-1-3}{x^2-x+1}dx\\\\
=x-\frac{1}{3}ln|x+1|+\frac{1}{6}ln|x^2-x+1|\\\\+\frac{1}{6}\int\frac{-3}{x^2-x+1}dx\\\\
=x-\frac{1}{3}ln|x+1|+\frac{1}{6}ln|x^2-x+1|\\\\-\frac{1}{2}\int\frac{1}{x^2-x+1}dx\\\\
=x-\frac{1}{3}ln|x+1|+\frac{1}{6}ln|x^2-x+1|\\\\-\frac{1}{2}\int\frac{1}{(x-\frac{1}{2})^2+(\frac{\sqrt{3}}{2}})^2dx\\\\
=x-\frac{1}{3}ln|x+1|+\frac{1}{6}ln|x^2-x+1|\\\\-\frac{1}{\sqrt{3}}tan^{-1}(\frac{2x-1}{\sqrt{3}})+C\\\\


\end{array}\]

Answer 24

Practicing latex ? ;)

Answer 25

practicing yes, but trying moreso to make my homework legible :)