So, by using the formula, we can solve the integral in the previous part.

[teng125]

does aanybody knows how to do this: int (x-2) / (x^2 +2x +3 )

pls help...

 

[gaganpreetsingh]

does aanybody knows how to do this: int (x-2) / (x^2 +2x +3 )

write the numerator in the form 1/2[(2x+2)-6]
now you get on dividing 1/2 { (2x+2)/(x^2+2x+3) - 6/(x^2 +2x +3)}
for the first part you cand substitute x^2 +2x +3 as t and it will simplify and for the second part make factors (if possible) and use partial fractions or make a perfect square.

PS this is supposed to be a calculus problem.

 

[teng125]

the answer for second part is arctan (x+1) / sqr root 2...
may i know how do u get it??

 

[Gamma]

In the second part, in the denominator, you have x^2 +2x+3.

Write it as (x+1)^2 + 2.

use the fact that d/dx (arctan(x/a) = 1/(x^2 + a^2)

 

[VietDao29]

In the second part, in the denominator, you have x^2 +2x+3.

Write it as (x+1)^2 + 2.

use the fact that d/dx (arctan(x/a) = 1/(x^2 + a^2)

Are you sure that: d/dx (arctan(x/a) = 1/(x^2 + a^2)? Shouldn't it read:
$$\frac{1}{a} \ \frac{d}{dx} \arctan \left( \frac{x}{a} \right) = \frac{1}{x ^ 2 + a ^ 2}$$?
$$\int \frac{x - 2}{x ^ 2 + 2x + 3} dx = \frac{1}{2} \ \int \frac{2x - 4}{x ^ 2 + 2x + 3} dx = \frac{1}{2} \ \int \frac{2x + 2 - 6}{x ^ 2 + 2x + 3} dx$$
$$= \frac{1}{2} \left( \int \frac{2x + 2}{x ^ 2 + 2x + 3} dx \ - \ \int \frac{6}{(x + 1) ^ 2 + 2} dx \right) = ...$$
This is just what others have shown you typed in LaTeX.
Can you go from here?
Note that:
$$\int \frac{dx}{x ^ 2 + a ^ 2} = \frac{1}{a} \arctan \left( \frac{x}{a} \right) + C$$