? ln x dx Top symbol and lower symbol on the intergral [ ?, 1] ? Top 1 bottom. S

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Explanation / Answer

Let u = [ln(x)]^2 and dv = dx ==> du = 2ln(x)/x and v = x. Then, we have: ? [ln(x)]^2 dx => uv - ? v du = x*[ln(x)]^2 - ? x[2ln(x)/x] dx = x*[ln(x)]^2 - 2 ? ln(x) dx. Then, integrating by parts again with u = ln(x) and dv = dx ==> du = 1/x dx and v = x: x*[ln(x)]^2 - 2 ? ln(x) dx => x*[ln(x)]^2 - 2(uv - ? v du) = x*[ln(x)]^2 - 2[x*ln(x) - ? (x)(1/x) dx] = x*[ln(x)]^2 - 2x*ln(x) + 2 ? dx = x*[ln(x)]^2 - 2x*ln(x) + 2x + C. Thus: ? [ln(x)]^2 dx (from x=1 to e) = x*[ln(x)]^2 - 2x*ln(x) + 2x (evaluated from x=1 to e) = [e(1)^2 - 2e(1) + 2e] - [1(0)^2 - 2(1)(0) + 2(1)] = e - 2. I hope this helps!

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