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Integral Calculus

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11 Jan 2012 13:56:55 IST

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Integrate

Mathematics

(1/x^2+x)dx

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Arjun Virmani

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Joined: 1 Jun 2009

Posts: 1294

11 Jan 2012 20:33:06 IST

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let I = 1/ (x^2 + x)

taking den.

(x^2 + x) = (x^2 + 1/2 . 2 . x + 1/4 - 1/4) = ( x + 1/2)^2 - 1/4

let t = ( x + 1/2) then dx = dt........

and ( x + 1/2)^2 - 1/4 = t^2 - 1/4 = (t - 1/2)(t + 1/2)...........

taking num

1 = (t + 1/2) - (t - 1/2)................

taking num / den.

I = [ (t + 1/2) - (t - 1/2) / (t - 1/2)(t + 1/2) ] dt

= [ 1 / (t - 1/2) - 1/ (t + 1/2) ] dt

= ln (t - 1/2) - ln (t + 1/2) + c

= ln [ (t - 1/2) / (t + 1/2) ] + c

substituting back t = ( x + 1/2)

I = ln [ (( x + 1/2) - 1/2) / (( x + 1/2) + 1/2) ] + c

= ln ( x / x + 1/4 ) +c

Abhishek Bandiya

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Joined: 7 Jul 2011

Posts: 59

13 Jan 2012 00:51:22 IST

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HI IF QUESTION IS 1/(X^2 + X)dxThen solution----Let A= 1/(X^2 +X) Taking x common from denominator A = 1/X(X +1)Adding x and subtracting x from numeratorA=( 1+X ?X)/X(1+X) A= [ (1+X)-X]/X(1+X) A = (1+X)/X(1+X) ?(X)/X(1+X)A= 1/X -1/(1+X) INTEGRATING I WITH RESPECT TO X Adx = 1/X dx ?(1)/(1+X)dx Adx = ln(X) ? ln(1+X) +C Adx = ln [ X/(1+X)] +C 1/X^2+X dx = ln [ X/(1+X)]+C

Abhishek Bandiya

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Joined: 7 Jul 2011

Posts: 59

13 Jan 2012 00:56:00 IST

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HI IF QUESTION IS 1/(X^2 + X)dx

Then solution----

Let A= 1/(X^2 +X)

Taking x common from denominator

A = 1/X(X +1)

Adding x and subtracting x from numerator

A=( 1+X –X)/X(1+X)

A= [ (1+X) -X] /X(1+X)

A = (1+X)/X(1+X) –(X)/X(1+X)

A= 1/X -1/(1+X)

INTEGRATION I WITH RESPECT TO X

Adx = 1/X dx –(1)/(1+X)dx

Adx = ln(X) – ln(1+X) +C

Adx = ln [ X/(1+X)] +C

1/X^2+X dx = ln [ X/(1+X)] +C

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