Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

# Integrating problem?

How do I integrate:

1/(2e^x + 3) dx?

Update:

Can you also include the steps?

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• J D
Lv 5

∫1/(2e^x + 3) dx

let t = e^x; dt = e^x dx

1/e^x dt = dx = 1/t dt

∫1/t(2t + 3) dt

1/t(2t + 3) = A/t + B/(2t+3)

1 = A(2t + 3) + Bt

at t = 0; A = 1/3

at t = 1; B = -2/3

So ∫1/t(2t + 3) dt =

1/3∫1/t dt + -2/3∫1/(2t + 3) dt

let w = 2t + 3; dw = 2 dt

1/3∫1/t dt + -2/6∫1/w dw

1/3∫1/t dt + -1/3∫1/w dw

=1/3 ln|t| -1/3 ln|w| + c

=1/3 ln|t| -1/3 ln|2t + 3| + c

=1/3 ln|e^x| -1/3 ln|2e^x + 3| + c

=x/3 -1/3 ln|2e^x + 3| + c

• 1 decade ago

Check wolfram or a calculator:

the integral is (1/3)(x - ln( 2e^x + 3 ) )

• 1 decade ago

Let y = 2e^x + 3

=> dy = 2e^x dx = (y-3) dx

int 1/(2e^x + 3) dx

= int 1/y * 1/(y-3) dy

= int 1/3 [1/(y-3) - 1/y] dy

= 1/3 [ln (y-3) - ln y] + c

= 1/3 ln [(y-3)/y] + c

= 1/3 ln [ (2e^x) / (2e^x +3) ] + c

• 3 years ago

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