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Evaluate the integral integral (1-x)/(1+2x) dx using the substitution method.

Mathematics

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Answer

34ln⁡∣1+2x∣−12x+C\frac{3}{4} \ln|1+2x| - \frac{1}{2}x + C

Step-by-step solution

6 steps

To solve the integral ∫1−x1+2xdx\int \frac{1-x}{1+2x} dx using the substitution method:

  1. Choose a substitution.

    Let u=1+2xu = 1+2x.

  2. Find dudu in terms of dxdx and express xx in terms of uu.

    Differentiating uu with respect to xx:

    dudx=2\frac{du}{dx} = 2

    So, dx=12dudx = \frac{1}{2} du.
    From u=1+2xu = 1+2x, we can express xx:

    u−1=2xu-1 = 2x x=u−12x = \frac{u-1}{2}

  3. Substitute xx into the numerator 1−x1-x.

    1−x=1−(u−12)1-x = 1 - \left(\frac{u-1}{2}\right) 1−x=2−(u−1)21-x = \frac{2 - (u-1)}{2} 1−x=2−u+121-x = \frac{2 - u + 1}{2} 1−x=3−u21-x = \frac{3-u}{2}

  4. Rewrite the integral in terms of uu.

    ∫1−x1+2xdx=∫3−u2u(12du)\int \frac{1-x}{1+2x} dx = \int \frac{\frac{3-u}{2}}{u} \left(\frac{1}{2} du\right) =∫3−u4udu= \int \frac{3-u}{4u} du =14∫(3u−uu)du= \frac{1}{4} \int \left(\frac{3}{u} - \frac{u}{u}\right) du =14∫(3u−1)du= \frac{1}{4} \int \left(\frac{3}{u} - 1\right) du

  5. Integrate with respect to uu.

    =14(3ln⁡∣u∣−u)+C= \frac{1}{4} \left(3 \ln|u| - u\right) + C

  6. Substitute back u=1+2xu = 1+2x.

    =14(3ln⁡∣1+2x∣−(1+2x))+C= \frac{1}{4} \left(3 \ln|1+2x| - (1+2x)\right) + C =34ln⁡∣1+2x∣−14(1+2x)+C= \frac{3}{4} \ln|1+2x| - \frac{1}{4}(1+2x) + C =34ln⁡∣1+2x∣−14−12x+C= \frac{3}{4} \ln|1+2x| - \frac{1}{4} - \frac{1}{2}x + C

    The constant −14-\frac{1}{4} can be absorbed into the arbitrary constant CC.

    The final answer is:

    34ln⁡∣1+2x∣−12x+C\frac{3}{4} \ln|1+2x| - \frac{1}{2}x + C
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