1. Integration Question

Hi
I need help on the following questions:

1) $\int \frac{x^2}{6x^2-7x+2}$
This what i have done:

do long division i get: $1 - \frac{5x^2-7x+2}{6x^2-7x+2}$

$\int 1dx \int \frac{5x^2-7x+2}{6x^2-7x+2}$

do partial fractions we get:

$\frac{-3}{2x-1} - \frac{1}{(3x-2)}$

final solution i got is:
$x - 3ln|2x-1| - ln|3x-2|$

2) $\int \frac{x-1}{3x^2+4x+2}$

$\frac{1}{6} \int \frac{6x+4}{3x^2+4x-2} + \int \frac{1}{3x^2+4x-2}$

$u=3x^2+4x+2 du=6x+4$

$\frac{1}{6} \int \frac{du}{u} + \int \frac{1}{3x^2+4x+2}$

$\frac{1}{6}ln|3x^2+4x+2| + \int \frac{1}{(3x+\frac{2}{\sqrt{3}}+\frac{14}{9})}$

final solution i got is:
$\frac{1}{6}ln|3x^2+4x+2| + \frac{1}{9}arcttan(3x+ \frac{2}{\sqrt{3}}) +c$

3) $\int \frac{3}{(x-+1)^2)(2x-3)}$
This what i have done:

split into partial fractions, after i get:

$A=\frac{3}{5}$ $B=\frac{6}{25}$ $C=\frac{-12}{25}
$

$\int \frac{A}{(x-1)^2} + \frac{B}{(x+1)} + \frac{C}{(2x-3)} =
\frac{3}{5(x-1)^2} + \frac{6}{25(x+1)} + \frac{-12}{25(2x-3)}$

final solution i got is:

$\frac{3}{5(x-1)^2} + \frac{6}{25}ln|x+1| - \frac{24}{25}ln|2x-3|$

P.S

2. Originally Posted by Paymemoney
Hi
I need help on the following questions:

1) $\int \frac{x^2}{6x^2-7x+2}$
This what i have done:

do long division i get: $1 - \frac{5x^2-7x+2}{6x^2-7x+2}$
Hi

Using long division you should find a numerator whose degree is lower than the denominator one

$\frac{x^2}{6x^2-7x+2} = \frac16 + \frac{ax+b}{6x^2-7x+2}$

3. Originally Posted by Paymemoney
Hi
I need help on the following questions:

1) $\int \frac{x^2}{6x^2-7x+2}$
This what i have done:

do long division i get: $1 - \frac{5x^2-7x+2}{6x^2-7x+2}$

$\int 1dx \int \frac{5x^2-7x+2}{6x^2-7x+2}$

do partial fractions we get:

$\frac{-3}{2x-1} - \frac{1}{(3x-2)}$

final solution i got is:
$x - 3ln|2x-1| - ln|3x-2|$

2) $\int \frac{x-1}{3x^2+4x+2}$

$\frac{1}{6} \int \frac{6x+4}{3x^2+4x-2} + \int \frac{1}{3x^2+4x-2}$

$u=3x^2+4x+2 du=6x+4$

$\frac{1}{6} \int \frac{du}{u} + \int \frac{1}{3x^2+4x+2}$

$\frac{1}{6}ln|3x^2+4x+2| + \int \frac{1}{(3x+\frac{2}{\sqrt{3}}+\frac{14}{9})}$

final solution i got is:
$\frac{1}{6}ln|3x^2+4x+2| + \frac{1}{9}arcttan(3x+ \frac{2}{\sqrt{3}}) +c$

3) $\int \frac{3}{(x-+1)^2)(2x-3)}$
This what i have done:

split into partial fractions, after i get:

$A=\frac{3}{5}$ $B=\frac{6}{25}$ $C=\frac{-12}{25}
$

$\int \frac{A}{(x-1)^2} + \frac{B}{(x+1)} + \frac{C}{(2x-3)} =
\frac{3}{5(x-1)^2} + \frac{6}{25(x+1)} + \frac{-12}{25(2x-3)}$

final solution i got is:

$\frac{3}{5(x-1)^2} + \frac{6}{25}ln|x+1| - \frac{24}{25}ln|2x-3|$

P.S

$\frac{x^2}{6x^2 - 7x + 2} = \frac{1}{6} + \frac{7x - 2}{6(6x^2 - 7x + 2)}$.

So $\int{\left(\frac{x^2}{6x^2 - 7x + 2}\right)\,dx} = \int{\left[\frac{1}{6} + \frac{7x - 2}{6(6x^2 - 7x + 2)}\right]\,dx}$

$= \int{\left(\frac{1}{6}\right)\,dx} + \frac{1}{6}\int{\left(\frac{7x - 2}{6x^2 - 7x + 2}\right)\,dx}$

$= \frac{x}{6} + \frac{7}{6}\int{\left(\frac{x - \frac{2}{7}}{6x^2 - 7x + 2}\right)\,dx}$

$= \frac{x}{6} + \frac{7}{72}\int{\left(\frac{12x - \frac{24}{7}}{6x^2 - 7x + 2}\right)\,dx}$

$= \frac{x}{6} + \frac{7}{72}\int{\left(\frac{12x - 7 + \frac{25}{7}}{6x^2 - 7x + 2}\right)\,dx}$

$= \frac{x}{6} + \frac{7}{72}\int{\left(\frac{12x - 7}{6x^2 - 7x + 2}\right)\,dx} + \frac{7}{72}\int{\left(\frac{\frac{25}{7}}{6x^2 - 7x + 2}\right)\,dx}$

$= \frac{x}{6} + \frac{7}{72}\int{\left(\frac{1}{u}\right)\,du} + \frac{25}{72}\int{\left(\frac{1}{6x^2 - 7x + 2}\right)\,dx}$ after making the substitution $u = 6x^2 - 7x + 2$

$= \frac{x}{6} + \frac{7\ln{|u|}}{72} + \frac{25}{432}\int{\left(\frac{1}{x^2 - \frac{7}{6}x + \frac{1}{3}}\right)\,dx}$

$= \frac{x}{6} + \frac{7\ln{|6x^2 - 7x + 2|}}{72} + \frac{25}{432}\int{\left[\frac{1}{x^2 - \frac{7}{6}x + \left(\frac{7}{12}\right)^2 - \left(\frac{7}{12}\right)^2 + \frac{1}{3}}\right]\,dx}$

$= \frac{x}{6} + \frac{7\ln{|6x^2 - 7x + 2|}}{72} + \frac{25}{432}\int{\left[\frac{1}{\left(x - \frac{7}{12}\right)^2 - \frac{1}{144}}\right]\,dx}$

Now making the substitution $x - \frac{7}{12} = \frac{1}{12}\cosh{t}$ so that $dx = \frac{1}{12}\sinh{t}\,dt$ this becomes

$= \frac{x}{6} + \frac{7\ln{|6x^2 - 7x + 2|}}{72} + \frac{25}{432}\int{\left[\frac{1}{\left(\frac{1}{12}\cosh{t}\right)^2 - \frac{1}{144}}\right]\,\frac{1}{12}\sinh{t}\,dt}$

$= \frac{x}{6} + \frac{7\ln{|6x^2 - 7x + 2|}}{72} + \frac{25}{432}\int{\left[\frac{144\sinh{t}}{12(\cosh^2{t} - 1)}\right]\,dt}$

$= \frac{x}{6} + \frac{7\ln{|6x^2 - 7x + 2|}}{72} + \frac{25}{432}\int{\left[\frac{12\sinh{t}}{\sinh{t}}\right]\,dt}$

$= \frac{x}{6} + \frac{7\ln{|6x^2 - 7x + 2|}}{72} + \frac{25}{36}\int{(1)\,dt}$

$= \frac{x}{6} + \frac{7\ln{|6x^2 - 7x + 2|}}{72} + \frac{25t}{36} + C$

$= \frac{x}{6} + \frac{7\ln{|6x^2 - 7x + 2|}}{72} + \frac{25\cosh^{-1}(12x - 7)}{36} + C$

$= \frac{12x + 7\ln{|6x^2 - 7x + 2|} + 50\cosh^{-1}(12x - 7)}{72} + C$.

4. $\int \frac{x^2dx}{6x^2-7z+2}$ has partial fractions
$\int \frac{4}{3(3x-2)}dx - \int\frac{1}{2(2x-1)}dx + \int 1/6dx$ and thus solution:
$x/6 - \frac{ln(x-1/2)}{4} + \frac{4ln(x-2/3)}{9}$.
Note that your partial fractions can't be correct because if you try to work backwards from your result you've lost the $x^2$ term in the numerator

5. Originally Posted by Paymemoney
Hi
I need help on the following questions:

1) $\int \frac{x^2}{6x^2-7x+2}$
This what i have done:

do long division i get: $1 - \frac{5x^2-7x+2}{6x^2-7x+2}$
The other posters have done a great job in completing this problem. I would just like to shed some light on long division...I absolutely, and I mean, absolutely hate it.

So what I do is find factors such that we can reduce the numerator.

$\frac{x^2}{6x^2-7x+2}$

$= \frac{6}{6}[ \frac{x^2 - \frac{7}{6} x + \frac{7}{6} x + \frac{2}{6} - \frac{2}{6} }{6x^2-7x+2}]$

$= \frac{1}{6} [ \frac{ 6x^2 - 7x + 2 +[7x-2] }{6x^2-7x+2}]$

$= \frac{1}{6} [ 1 + \frac{ 7x -2 }{6x^2-7x+2}]$

Which we can now decompose by partial fractions.

No long division required

6. Originally Posted by GeoC
$\int \frac{x^2dx}{6x^2-7z+2}$ has partial fractions
$\int \frac{4}{3(3x-2)}dx - \int\frac{1}{2(2x-1)}dx + \int 1/6dx$ and thus solution:
$x/6 - \frac{ln(x-1/2)}{4} + \frac{4ln(x-2/3)}{9}$.
how did you get this solution i always get $\frac{-3}{2x-1} + \frac{8}{3x-2}$from doing partial fractions.

7. Originally Posted by Paymemoney
how did you get this solution i always get $\frac{-3}{2x-1} + \frac{8}{3x-2}$from doing partial fractions.
$\frac{x^2}{6x^2 - 7x + 2} = \frac{1}{6} + \frac{7x - 2}{6(6x^2 - 7x + 2)}$

$= \frac{1}{6} + \frac{1}{6}\left[\frac{7x - 2}{(3x- 2)(2x - 1)}\right]$

Now using partial fractions:

$\frac{A}{3x - 2} + \frac{B}{2x - 1} = \frac{7x - 2}{(3x - 2)(2x - 1)}$

$\frac{A(2x - 1) + B(3x - 2)}{(3x - 2)(2x - 1)} = \frac{7x - 2}{(3x - 2)(2x - 1)}$

$A(2x - 1) + B(3x - 2) = 7x - 2$

$2Ax - A + 3Bx - 2B = 7x - 2$

$(2A + 3B)x - (A + 2B) = 7x - 2$.

So now your system of equations is

$2A + 3B = 7$

$A + 2B = 2$.

Multiply the second equation by 2...

$2A + 3B = 7$

$2A + 4B = 4$.

Subtract the second equation from the first...

$(2A + 3B) - (2A + 4B) = 7 - 4$

$-B = 3$

$B = -3$.

Substituting back into equation 2...

$A + 2B = 2$

$A + 2(-3) = 2$

$A - 6 = 2$

$A = 8$.

So $\frac{7x - 2}{(3x - 2)(2x - 1)} = \frac{8}{3x - 2} - \frac{3}{2x - 1}$

and therefore

$\frac{1}{6} + \frac{1}{6}\left[\frac{7x - 2}{(3x- 2)(2x - 1)}\right] = \frac{1}{6} + \frac{1}{6}\left(\frac{8}{3x - 2} - \frac{3}{2x - 1}\right)$

$= \frac{1}{6} + \frac{4}{3(3x - 2)} - \frac{1}{2(2x - 1)}$.

So that means

$\int{\frac{x^2}{6x^2 - 7x + 2}\,dx} = \int{\frac{1}{6}\,dx} + \frac{4}{3}\int{\frac{1}{3x - 2}\,dx} - \frac{1}{2}\int{\frac{1}{2x - 1}\,dx}$

$= \frac{x}{6} + \frac{4}{9}\int{\frac{3}{3x - 2}\,dx} - \frac{1}{4}\int{\frac{2}{2x - 1}\,dx}$

$= \frac{x}{6} + \frac{4\ln{|3x - 2|}}{9} - \frac{\ln{|2x - 1|}}{4} + C$

$= \frac{6x + 16\ln{|3x - 2|} - 9\ln{|2x - 1|}}{36} + C$.

8. Subtract the second equation from the first...

$(2A + 3B) - (2A + 4B) = 7 - 4$

$-B = 3$

$B = 3$.

is this meant to be -3 not 3

9. this is how i did it, i don't understand how my way is incorrect.

$\frac{A}{2x-1} + \frac{B}{3x-2}$

$A(3x-2) + B(2x-1) = 7x-2$

3A + 2B = 7 (1
-2A - B = -2 (2

times 2

-4A - B = -4

minus 1) from 2)

-A = 3
A = -3

therefore B = $\frac{16}{2} = 8$

so i get

$\frac{-3}{(2x-1)} + \frac{8}{(3x-2)}$

10. Originally Posted by Paymemoney
this is how i did it, i don't understand how my way is incorrect.

$\frac{A}{2x-1} + \frac{B}{3x-2}$

$A(3x-2) + B(2x-1) = 7x-2$

3A + 2B = 7 (1
-2A - B = -2 (2

times 2

-4A - B = -4

minus 1) from 2)

-A = 3
A = -3

therefore B = \frac{16}{2} = 8

so i get

$\frac{-3}{(2x-1)} + \frac{8}{(3x-2)}$

11. Originally Posted by Paymemoney
is this meant to be -3 not 3
Yes it is - edited.

12. Originally Posted by Paymemoney
this is how i did it, i don't understand how my way is incorrect.

$\frac{A}{2x-1} + \frac{B}{3x-2}$

$A(3x-2) + B(2x-1) = 7x-2$

3A + 2B = 7 (1
-2A - B = -2 (2

times 2

-4A - 2B = -4

minus 1) from 2) Since the signs are different for B here, you need to add the equations.
From here you should have gotten:

$-A = 3$ which means $A = -3$.

Sub into 2)

$-2(-3) - B = -2$

$6 - B = -2$

$B = 8$.

13. Just to give a different method

From

$\frac{A}{3x - 2} + \frac{B}{2x - 1} = \frac{7x - 2}{(3x - 2)(2x - 1)}$

I multiply both sides by $3x - 2$

$A + \frac{B(3x-2)}{2x - 1} = \frac{7x - 2}{2x - 1}$

And I set $x = \frac23$

$A = \frac{7 \frac23 - 2}{2 \frac23 - 1} = \frac{\frac83}{\frac13} = 8$

I do the same for $2x - 1$

$\frac{A(2x-1)}{3x - 2} + B = \frac{7x - 2}{3x - 2}$

And I set $x = \frac12$

$B = \frac{7 \frac12 - 2}{3 \frac12 - 2} = \frac{\frac32}{-\frac12} = -3$

14. Can anyone answer my other two question??

15. Originally Posted by Paymemoney
2) $\int \frac{x-1}{3x^2+4x+2}$

$\frac{1}{6} \int \frac{6x+4}{3x^2+4x-2} + \int \frac{1}{3x^2+4x-2}$

$u=3x^2+4x+2 du=6x+4$

$\frac{1}{6} \int \frac{du}{u} + \int \frac{1}{3x^2+4x+2}$

$\frac{1}{6}ln|3x^2+4x+2| + \int \frac{1}{(3x+\frac{2}{\sqrt{3}}+\frac{14}{9})}$

final solution i got is:
$\frac{1}{6}ln|3x^2+4x+2| + \frac{1}{9}arcttan(3x+ \frac{2}{\sqrt{3}}) +c$
$\int {\frac{x-1}{3x^2+4x+2}\,dx} = \frac{1}{6}\int{\frac{6x - 6}{3x^2 + 4x + 2}\,dx}$

$= \frac{1}{6}\int{\frac{6x + 4}{3x^2 + 4x + 2}\,dx} - \frac{1}{6}\int{\frac{10}{3x^2 + 4x + 2}\,dx}$

$= \frac{1}{6}\int{\frac{1}{u}\,du} - \frac{5}{9}\int{\frac{1}{x^2 + \frac{4}{3}x + \frac{2}{3}}\,dx}$ after making the substitution $u = 3x^2 + 4x + 2$

$= \frac{\ln|u|}{6} - \frac{5}{9}\int{\frac{1}{x^2 + \frac{4}{3}x + \left(\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2 + 2}\,dx}$

$= \frac{\ln{|3x^2 + 4x + 2|}}{6} - \frac{5}{9}\int{\frac{1}{\left(x + \frac{2}{3}\right)^2 + \frac{14}{9}}\,dx}$

Now make the substitution $x + \frac{2}{3} = \frac{\sqrt{14}}{3}\tan{\theta}$ so that $dx = \frac{\sqrt{14}}{3}\sec^2{\theta}\,d\theta$ and the integral becomes

$= \frac{\ln{|3x^2 + 4x + 2|}}{6} - \frac{5}{9}\int{\frac{1}{\frac{14}{9}\tan^2{\theta } + \frac{14}{9}}\,\frac{\sqrt{14}}{9}\sec^2{\theta}\, d\theta}$

$= \frac{\ln{|3x^2 + 4x + 2|}}{6} - \frac{5}{9}\int{\frac{\sqrt{14}\sec^2{\theta}}{14( \tan^2{\theta} + 1)}\,d\theta}$

$= \frac{\ln{|3x^2 + 4x + 2|}}{6} - \frac{5\sqrt{14}}{126}\int{\frac{\sec^2{\theta}}{\ sec^2{\theta}}\,d\theta}$

$= \frac{\ln{|3x^2 + 4x + 2|}}{6} - \frac{5\sqrt{14}}{126}\int{1\,d\theta}$

$= \frac{\ln{|3x^2 + 4x + 2|}}{6} - \frac{5\sqrt{14}\,\theta}{126} + C$

$= \frac{\ln{|3x^2 + 4x + 2|}}{6} - \frac{5\sqrt{14}\left[\frac{3}{\sqrt{14}}\arctan{\left(x + \frac{2}{3}\right)}\right]}{126} + C$

$= \frac{\ln{|3x^2 + 4x + 2|}}{6} - \frac{5\arctan{\left(x + \frac{2}{3}\right)}}{42} + C$

$= \frac{7\ln{|3x^2 + 4x + 2|} - 5\arctan{\left(x + \frac{2}{3}\right)}}{42} + C$.