Integration help pleaseee

[ninbam1k]

hey guys, help me out with this one, integrate: 2+(tanx)^2 with respect to x.

thanks peeps

[brightsky]

Evaluating ,

Remember

So we need to evaluate

So the whole integral is

[ninbam1k]

oh shit, i got inte of (secx)^2 = tanx, thanks mate

[yawho]

any idea anyone?
integrate (2x^5-8x^3+6x)/(x^6-6x^4+12x^2-8) dx

[brightsky]

Expand into partial fractions (the denominator is factorisable) and antidiff it term by term.

[yawho]

please show how

[brightsky]

(2x^5-8x^3+6x)/(x^6-6x^4+12x^2-8) = 2x(x-1)(x+1)(x^2 - 3)/(x^2 - 2)^3 = Ax/(x^2 - 2) + Bx/(x^2 - 2)^3
Ax(x^2 - 2)^2 + Bx = 2x^5-8x^3+6x
Ax(x^4 - 4x^2 + 4) + Bx = 2x^5 - 8x^3 + 6x
Ax^5 - 4Ax^3 + (4A + B)x = 2x^5 - 8x^3 + 6x
A = 2, 4A + B = 6, so B = -2.
so the integral becomes
int 2x/(x^2 - 2) dx - int 2x/(x^2 - 2)^3 dx
for the first integral,
let u = x^2 - 2, du = 2x dx
so we have int 1/u du = ln|u| + C = ln|x^2 - 2| + C
for the second integral,
let u = x^2 - 2, du = 2x dx
so we have int 1/u^3 du = u^(-2)/(-2) + C = -1/2(x^2 - 2) + C
so combining them together we get the answer:
= ln|x^2 - 2| + 1/2(x^2 - 2) + C

[yawho]

how do you know what are the partial fractions?

[ninbam1k]

sorry, just need help on this last one, i find it so difficult to figure out. Integrate e^2x / sqrt(e^x  +  1).

[TrueTears]

Rest is easy

[ninbam1k]

thanks simon you gangster!

[yawho]

(2x^5-8x^3+6x)/(x^6-6x^4+12x^2-8) = 2x(x-1)(x+1)(x^2 - 3)/(x^2 - 2)^3 = Ax/(x^2 - 2) + Bx/(x^2 - 2)^3
Ax(x^2 - 2)^2 + Bx = 2x^5-8x^3+6x
Ax(x^4 - 4x^2 + 4) + Bx = 2x^5 - 8x^3 + 6x
Ax^5 - 4Ax^3 + (4A + B)x = 2x^5 - 8x^3 + 6x
A = 2, 4A + B = 6, so B = -2.
so the integral becomes
int 2x/(x^2 - 2) dx - int 2x/(x^2 - 2)^3 dx
for the first integral,
let u = x^2 - 2, du = 2x dx
so we have int 1/u du = ln|u| + C = ln|x^2 - 2| + C
for the second integral,
let u = x^2 - 2, du = 2x dx
so we have int 1/u^3 du = u^(-2)/(-2) + C = -1/2(x^2 - 2) + C
so combining them together we get the answer:
= ln|x^2 - 2| + 1/2(x^2 - 2) + C

Can you/anyone explain the partial fractions part please.

[costa94]

(2x^5-8x^3+6x)/(x^6-6x^4+12x^2-8) = 2x(x-1)(x+1)(x^2 - 3)/(x^2 - 2)^3 = Ax/(x^2 - 2) + Bx/(x^2 - 2)^3
Ax(x^2 - 2)^2 + Bx = 2x^5-8x^3+6x
Ax(x^4 - 4x^2 + 4) + Bx = 2x^5 - 8x^3 + 6x
Ax^5 - 4Ax^3 + (4A + B)x = 2x^5 - 8x^3 + 6x
A = 2, 4A + B = 6, so B = -2.
so the integral becomes
int 2x/(x^2 - 2) dx - int 2x/(x^2 - 2)^3 dx
for the first integral,
let u = x^2 - 2, du = 2x dx
so we have int 1/u du = ln|u| + C = ln|x^2 - 2| + C
for the second integral,
let u = x^2 - 2, du = 2x dx
so we have int 1/u^3 du = u^(-2)/(-2) + C = -1/2(x^2 - 2) + C
so combining them together we get the answer:
= ln|x^2 - 2| + 1/2(x^2 - 2) + C

Can you/anyone explain the partial fractions part please.

factorise both top and bottom, then equate it equal to "Ax/(x^2 - 2) + Bx/(x^2 - 2)^3" (notice factors of the denominator on the whole fraction on left side are now factors of our two fractions on right).

from there make the "Ax/(x^2 - 2) + Bx/(x^2 - 2)^3" bit one single fraction, and from there you can solve the numerators of both equal to each other to find values of A and B, to therefore integrate the two partial fractions

(probably not the best worded explanation, but tired + bored after writing english oral for sac so leave me alone )

[yawho]

How did you arrive at the partial fractions Ax/(x^2 - 2) + Bx/(x^2 - 2)^3? This is what I did not understand.