Questions from Coroneos' 100 integrals (1 Viewer)

MC Squidge · May 5, 2009

Bro i cant do number 20, 50 and 56. help

jet · May 5, 2009

Type them out.

L · May 5, 2009

aah. the good old 100 integrals

MC Squidge · May 5, 2009

20. Integral of (x)/((x)^(1/2)+1) dx

50. Integral of (x^3)((4+x^2)^(-1/2)) dx

56. Integral of (Sec Squared X)/(Tan Squared x - 3Tanx +2) dx

Drongoski · May 6, 2009

$\100dpi Q20\\ \\ Let\ u=\sqrt{x}\ \ \therefore \ \ x=u^{2}\ \ \ \ dx=2udu\\ \\ \therefore \ \ \int \frac{x}{\sqrt{x}+1}\ dx\ \ =\ \ \int \frac{u^{2}\times 2udu}{u+1}\ \ =\ \ \int \frac{2u^{3}}{u+1}\ du\\ \\ \ \ =\ \ 2\int \left [u^{2}-u+1\ -\ \frac{1}{u+1} \right ]\ du\\ \\ \ \ =\ \ \frac{2}{3}u^{3}\ -\ u^{2}\ +\ 2u\ -\ 2ln\left |u+1 \right |\ +\ C\\ \\ You\ then\ unsubstitute\ back\ the\ x.\\ \\ \\ Q50\\ \\ \\ Let\ x=2tan\theta \ \ \therefore \ \ tan\theta =\frac{x}{2}\ \ \ dx=2sec^{2}\theta d\theta \ \ \ \sqrt{4+x^{2}}=\sqrt{4sec^{2}\theta }=2sec\theta \\ \\ \therefore \ \ \int \frac{x^{3}}{\sqrt{4+x^{2}}}dx\ \ =\ \ \int \frac{8tan^{2\theta }\times 2sec^{2}\theta d\theta} {2sec\theta } \ \ =\ \ 8\int tan^{2}\theta \times sec\theta tan\theta d\theta \\ \\ \ \ =\ \ 8\int (sec^{2}\theta -1)dsec\theta \ \ =\ \ 8\left (\frac{sec^{3}\theta }{3} -sec\theta \right )\ +\ C\ \ =\\ \\ \ \ =\ \ \frac{8}{3}sec\theta \left (tan^{2}\theta -3 \right )\ +\ C\ \ =\ \ \ \$

You then subst sec@ = sqr(4+x^2)/2 and tan@ = x/2

Drongoski · May 6, 2009

I had problem with LaTeX; so this is a continuation of previous.

$\100dpi \ \ =\ \ \frac{8}{24}\left [\sqrt{4+x^{2}}(x^{2}-12) \right ]\ +\ C\ \ =\ \ \frac{1}{3}\sqrt{4+x^{2}}(x^{^{2}}-12)\ +\ C\\ \\ \\ Q56\\ \\ \\ \int \frac{sec^{2}x}{tan^{2}x-3tanx+2}\ dx\ \ =\ \ \int \frac{1}{(tanx\ -\ 2)(tanx\ -\ 1)}\ sec^{2}x\ dx\\ \\ \\ \ =\ \ \int \left [\frac{1}{tanx-2}\ -\ \frac{1}{tanx-1} \right ]\ dtanx\ \ =\ \ ln\left |tanx-2 \right |-ln\left |tanx-1 \right |\ +\ C\\ \\ \\ \ \ =\ \ ln\left |\frac{tanx-2}{tanx-1} \right |\ +\ C$

Have no time to go thru my solutions; plz correct any errors found.

Kendall_Gilroy · May 6, 2009

L said:
aah. the good old 100 integrals

lol 100 aboriginals

cutemouse · May 6, 2009

I did them all... took me a god damn week to get the answers to match with the back of the book. But got all 100 done =D

gurmies · May 6, 2009

jm01 said:
I did them all... took me a god damn week to get the answers to match with the back of the book. But got all 100 done =D

JEEZ. Where did you get them from?

EDIT: Just realised I have the book that they are in. So I begin...

cutemouse · May 6, 2009

Yeah it's good practice if you get them to the answer at the back... My teacher was impressed apparently that I did that.

It is annoying doing it though...

MC Squidge · May 6, 2009

thanks for the help bros

i got some more i cannot do

63. Integral of (9+x^2)^(1/2) dx
64. Integral of x(9+x^2)^(1/2) dx
70. Integral of (x^3+1)/(x^3-x) dx having trouble dividing this one
72. Integral of ln(x+(x^2-1)^(1/2)) dx
84. Integral of (7+x)/(1+x+x^2+x^3) dx

Drongoski · May 6, 2009

Q70

$\100dpi \int \frac{x^{3}+1}{x^{3}-x}dx\ \ =\ \ \int \frac{x^{3}-x\ +\ x+1}{x^{3}-x}\ dx\\ \\ \\ \ =\ \int 1\ +\ \frac{x+1}{x(x-1)(x+1)}\ dx\ \ =\ \ \int 1+\frac{1}{x-1}-\frac{1}{x}\ \ dx\\ \\ \\ \ =\ \ x\ -\ ln\left |\frac{x-1}{x} \right |\ +\ C$

Timothy.Siu · May 6, 2009

63. Integral of (9+x^2)^(1/2) dx
64. Integral of x(9+x^2)^(1/2) dx
70. Integral of (x^3+1)/(x^3-x) dx having trouble dividing this one
72. Integral of ln(x+(x^2-1)^(1/2)) dx
84. Integral of (7+x)/(1+x+x^2+x^3) dx

63. just sub in 3tan @
64. same as 63
70. shown above
72. by parts i presume
84. i'm sure theres a better way but i think it = (7+x)((x-1)/(x^4-1)

Drongoski · May 6, 2009

Q63

$\100dpi Let\ x=3\ tan\theta \ \ \ \therefore \ dx=3sec^{2} \theta d\theta \ \ \ \ \sqrt{9+x^{2}}= 3sec\theta \\ \\ \\ \therefore \ \ I\ =\ \int \sqrt{9+x^{2}}\ dx\ \ =\ \ \int 3sec\theta \ \cdot \ 3sec^{2}\theta \ d\theta \\ \\ \\ \ =\ \ 9\int sec\theta \ dtan\theta \ \ =\ \ 9\left [sec\theta \ tan\theta \ -\ \int tan \theta \ \cdot \ sec\theta\ tan\theta \ d\theta \right ]\\ \\ \\ \ =\ \ 9\left [sec\theta \ tan\theta \ -\ \int (sec^{3}\theta -sec\theta )\ d\theta \right ]\ =\ 9\left [sec\theta \ tan\theta +\int sec\theta \ d \theta \right ]\ -\ I\\ \\ \\ \ \therefore \ \ 2I\ \ =\ \ 9\left [sec\theta \ tan\theta \ +\ ln\left |sec\theta +tan\theta \right | \right ]\\ \\ \ I\ \ =\ \ \frac{9}{2}\left [\ \ as\ above\ \ \right ]$

Drongoski · May 6, 2009

Q72

$\100dpi \int ln(x+\sqrt{x^{2}-1})\ dx\ \ =\ \ x\ ln(x+\sqrt{x^{2}-1})\ -\ \int x\ \frac{1}{\sqrt{x^{2}=1}}\ dx\ \\ \\ \\ \ =\ \ x\ ln(\cdots )\ -\ \sqrt{x^{2}-1}\ +\ C$

Trebla · May 6, 2009

MC Squidge said:
84. Integral of (7+x)/(1+x+x^2+x^3) dx

Note that: 1 + x + x² + x³ = (x + 1)(x² + 1)
then write (7 + x) / (1 + x + x² + x³) in the form a / (x + 1) + (bx + c) / (x² + 1) and it should be easy to integrate...

Drongoski · May 6, 2009

Q84

$\100dpi \int \frac{7+x}{1+x+x^{2}+x^{3}}\ dx\ \ =\ \ \int \frac{(1-x)(7+x)}{1-x^{4}}\ dx\\ \\ \\ \ =\ \ \int \frac{(1-x^{2})\ +\ 6(1-x)}{(1-x^{2})(1+x^{2})}\ dx\ \ =\ \ \int \left [\frac{4}{1+x^{2}} -\ \frac{3x}{1+x^{2}}\ +\ \frac{3}{1+x} \right ]\ dx\\ \\ \\ \ =\ 4\ tan^{-1}x\ -\ \frac{3} {2}\ ln(1+x^{2})\ +\ 3\ ln\left |1+x \right |\ +\ C$

Edit

Don't know if my solutions are correct! Because of difficulty typing out in LaTeX, I've left out many intermediate steps.

Edit-2

I've made a beautiful mistake. Above is amended solution; hopefully now correct.

gurmies · May 6, 2009

Drongoski said:
Q84

$\100dpi \int \frac{7+x}{1+x+x^{2}+x^{3}}\ dx\ \ =\ \ \int \frac{(1-x)(7+x)}{1-x^{4}}\ dx\\ \\ \\ \ =\ \ \int \frac{(1-x^{2})\ +\ 6(1-x)}{(1-x^{2})(1+x^{2})}\ dx\ \ =\ \ \int \left [\frac{1}{1+x^{2}} -\ \frac{6x}{1+x^{2}}\ +\ \frac{6}{1+x} \right ]\ dx\\ \\ \\ \ =\ tan^{-1}x\ -\ 3\ ln(1+x^{2})\ +\ 6\ ln\left |1+x \right |\ +\ C$

Beautiful

MC Squidge · May 7, 2009

thanks again

harism · May 7, 2009

gurmies said:
Beautiful

agreed.
very.

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