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saw5 [17]
2 years ago
13

Integrala x la a treia ori ln la a doua dx va rog

Mathematics
1 answer:
Studentka2010 [4]2 years ago
3 0

I don't speak Romanian, but the closest translation for this suggests you're trying to compute

\displaystyle \int x^3 \ln(x)^2 \, dx

Integrate by parts:

\displaystyle \int x^3 \ln(x)^2 \, dx = uv - \int v \, du

where

u = ln(x)²   ⇒   du = 2 ln(x)/x dx

dv = x³ dx   ⇒   v = 1/4 x⁴

\implies \displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \int x^3 \ln(x) \, dx

Integrate by parts again:

\displaystyle \int x^3 \ln(x) \, dx = u'v' - \int v' du'

where

u' = ln(x)   ⇒   du' = dx/x

dv' = x³ dx   ⇒   v' = 1/4 x⁴

\implies \displaystyle \int x^3 \ln(x) \, dx = \frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx

So, we have

\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \left(\frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx \right)

\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \int x^3 \, dx

\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \left(\frac14 x^4\right) + C

\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac1{32} x^4 + C

\boxed{\displaystyle \int x^3 \ln(x)^2 \, dx = \frac1{32} x^4 \left(8\ln(x)^2 - 4\ln(x) + 1\right) + C}

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Answer:

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Give me a comment if you want the explanation.

1.) \mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0

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2.) \mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0

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Step-by-step explanation:

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C

Step-by-step explanation:

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Step-by-step explanation:

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when you add 5y on both sides it cancels the 5y out, it's no solution.

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3 years ago
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