# Integrating problem?

How do I integrate:

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

Can you also include the steps?

### 5 Answers

- J DLv 51 decade agoBest Answer
∫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

- mathmaniacLv 61 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

- larrinagaLv 43 years ago
hi, The given answer is purely one answer of the Bernoulli differential equation, in case you're able to do the integrating ingredient approach, you will discover the final answer that way. in spite of the undeniable fact that, the assistance y(t) = t² satisfies the equation is extremely efficient. What you're able to do is think approximately the way this differentiates: d[t²] = 2t dt. This tells you that, in case you have been to multiply with the aid of by ability of a few consistent, say, c, it might stay after differentiation, and then cancel itself out interior the differential equation. that's: d[ct²] = 2ct dt. Or: d/dt[ct²] = 2ct. Substituting in: 2ct - [2 / t] ct² = 0. Factoring c and simplifying: c (2t - 2t) = 0. The bracket is obviously 0, subsequently, the cost of c as a multiplier is beside the point. ----- you need to no longer do this with y(t) = t² + c, with the aid of fact, on differentiation, the consistent disappears: d[t² + c] = 2t dt. so which you will no longer ingredient it out, and it is not beside the point. if reality be told, you will get: 2t - [2 / t] (t² + c) = 2t - 2t + 2c/t = 2c / t for sure, 2c / t ? 0 for each consistent c. ----- you could verify the respond by ability of utilising the integrating ingredient approach. that's a Bernoulli differential equation, with parameters: n = 0. P(t) = -2 / t. Q(t) = 0. The integrating ingredient is that if = e^(? P(t) dt). First, do the crucial: ? P(t) dt = ? -2dt / t = -2 ? dt / t = -2 ln(t) = ln(a million / t²). Then, calculate the integrating ingredient: IF = e^(? P(t) dt) = e^(ln(a million / t²)) = a million / t². Multiply the DE by ability of the IF: (dy/dt) / t² - [2 / t] [y / t²] = 0 / t². Simplify: (dy/dt) / t² - 2y / t³ = 0. The LHS is now an entire product derivative of y * IF: d/dt [y / t²] = 0. save on with integrals with appreciate to t: ? d/dt [y / t²] dt = ? 0 dt. Cancel derivative with crucial: y / t² = ? 0 dt. The crucial of 0 is in basic terms a relentless, say, c: y / t² = c. Multiply by ability of t²: y = ct². as expected. -----

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- Anonymous1 decade ago
ln (2e^x + 3) +c

instead of the brackets make a modulus sign

hope i helped