# Integration Question

#### Paymemoney

Hi
I need help on the following questions:

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

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

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

do partial fractions we get:

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

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

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

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

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

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

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

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

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

split into partial fractions, after i get:

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

final solution i got is:

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

P.S

Rapha

#### running-gag

Hi
I need help on the following questions:

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

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

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

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

#### Prove It

MHF Helper
Hi
I need help on the following questions:

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

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

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

do partial fractions we get:

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

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

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

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

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

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

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

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

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

split into partial fractions, after i get:

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

final solution i got is:

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

P.S

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Rapha and General

#### GeoC

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

#### AllanCuz

Hi
I need help on the following questions:

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

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

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

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

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

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

Which we can now decompose by partial fractions.

No long division required

Rapha

#### Paymemoney

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

#### Prove It

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

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

Now using partial fractions:

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

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

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

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

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

So now your system of equations is

$$\displaystyle 2A + 3B = 7$$

$$\displaystyle A + 2B = 2$$.

Multiply the second equation by 2...

$$\displaystyle 2A + 3B = 7$$

$$\displaystyle 2A + 4B = 4$$.

Subtract the second equation from the first...

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

$$\displaystyle -B = 3$$

$$\displaystyle B = -3$$.

Substituting back into equation 2...

$$\displaystyle A + 2B = 2$$

$$\displaystyle A + 2(-3) = 2$$

$$\displaystyle A - 6 = 2$$

$$\displaystyle A = 8$$.

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

and therefore

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

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

So that means

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

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

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

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

Last edited:

#### Paymemoney

Subtract the second equation from the first...

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

$$\displaystyle -B = 3$$

$$\displaystyle B = 3$$.

is this meant to be -3 not 3

#### Paymemoney

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

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

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

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

times 2

-4A - B = -4

minus 1) from 2)

-A = 3
A = -3

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

so i get

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

#### AllanCuz

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

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

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

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

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