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TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

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Question : 73 of 160

Marks: +1, -0

\int x^3 (\log x)^2 \, dx = x^4 [A (\log x)^2 + B (\log x) + C \log e] + K

$\frac{7}{24}$
$\frac{4}{25}$
$\frac{3}{14}$
$\frac{5}{32}$

Solution:

We have,

\int x^{3} (\log x)^{2} \, dx = x^{4}

[A (\log x)^{2} + B (\log x) + C \log e] + K

Now,

\int x^{3} (\log x)^{2} \, dx

= (\log x)^{2} \int x^{3} \, dx - \int \left( \frac{d}{dx} (\log x)^{2} \int x^{3} \, dx \right) dx

= (\log x)^{2} \cdot \frac{x^{4}}{4} - \int (2 \log x) \cdot \frac{1}{x} \cdot \frac{x^{4}}{4} \, dx

= (\log x)^{2} \left( \frac{x^{4}}{4} \right) - \frac{1}{2} \int \log x \cdot x^{3} \, dx

= (\log x)^{2} \left( \frac{x^{4}}{4} \right) - \frac{1}{2}

\left[\log x \int x^{3} \, dx - \int \left( \frac{d}{dx} \left(\log x \int x^{3} \, dx\right) \right) dx\right]

= (\log x)^{2} \left( \frac{x^{4}}{4} \right) - \frac{1}{2} \left[ \log x \left( \frac{x^{4}}{4} \right) - \int \frac{1}{x} \cdot \frac{x^{4}}{4} \, dx \right]

= (\log x)^{2} \left( \frac{x^{4}}{4} \right) - \frac{1}{2} \left[ \log x \left( \frac{x^{4}}{4} \right) - \frac{1}{4} \cdot \int x^{3} \, dx \right]

= (\log x)^{2} \left( \frac{x^{4}}{4} \right) - \frac{1}{2} \left[ \log x \left( \frac{x^{4}}{4} \right) - \frac{1}{4} \cdot \frac{x^{4}}{4} \right] + K

= x^{4} \left[ \frac{1}{4} (\log x)^{2} - \frac{1}{8} \log x + \frac{1}{32} \log e \right] + K

\therefore \;\; A = \frac{1}{4}, B = \frac{-1}{8}

and

C = \frac{1}{32}

\therefore A + B + C = \frac{1}{4} - \frac{1}{8} + \frac{1}{32} = \frac{8 - 4 + 1}{32} = \frac{5}{32}

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