Question

Вычислить интеграл ∫ 4xln^2xdx,интегрируя по частям. Заранее спасибо!

Answer 1

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Answer 2

Calculating the Integral ∫ 4xln^2xdx using Integration by Parts

To calculate the integral ∫ 4xln^2xdx using integration by parts, we can follow the formula:

∫ u dv = uv - ∫ v du

Let's assign u = ln^2x and dv = 4xdx. We can then find du and v:

du = (2lnx / x) dx v = ∫ 4xdx = 2x^2

Now, we can substitute these values into the integration by parts formula:

∫ 4xln^2xdx = u * v - ∫ v * du = ln^2x * 2x^2 - ∫ 2x^2 * (2lnx / x) dx

Simplifying further:

∫ 4xln^2xdx = 2x^2ln^2x - 2∫ xlnxdx

Now, we can integrate the remaining term using integration by parts again. Let's assign u = lnx and dv = xdx. We can find du and v:

du = (1 / x) dx v = ∫ xdx = (1/2)x^2

Substituting these values into the integration by parts formula:

∫ xlnxdx = u * v - ∫ v * du = lnx * (1/2)x^2 - ∫ (1/2)x^2 * (1 / x) dx = (1/2)x^2lnx - (1/2)∫ xdx = (1/2)x^2lnx - (1/4)x^2

Now, we can substitute this result back into the previous equation:

∫ 4xln^2xdx = 2x^2ln^2x - 2((1/2)x^2lnx - (1/4)x^2) = 2x^2ln^2x - x^2lnx + (1/2)x^2

Therefore, the integral ∫ 4xln^2xdx, integrated by parts, is equal to:

2x^2ln^2x - x^2lnx + (1/2)x^2 + C

where C is the constant of integration.

Please note that the above solution is derived using the integration by parts method.