( x)² x³ dx = \; {x^{4}}{32} f(x) + C f(x) =

Correct Answer is : 8(log⁡x)2−4log⁡x+18( x)² - 4 x + 18(logx)2−4logx+1

Correct Answer is : 8(log⁡x)2−4log⁡x+18( x)² - 4 x + 18(logx)2−4logx+1

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AP EAPCET 08-Jul-2022 Shift 2 Paper

Section: Mathematics

Question : 73 of 160

Marks: +1, -0

\int (\log x)^{2} x^{3} dx = \; \frac{x^{4}}{32} f(x) + C \Rightarrow f(x) =

Solution: 👈: Video Solution

I = \int (\log x)^2 x^3 dx

Using integration by parts,

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

= \frac{(\log x)^2 x^4}{4} - \int \frac{2 \log x}{x} \times \frac{x^4}{4} dx

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

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

(\log x) \int x^3 dx - \int \frac{d \log x}{dx} \int x^3 dx) dx]

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

= \frac{(\log x)^2 x^4}{4} - \frac{1}{8} (\log x) x^4 + \frac{1}{8} \times \frac{x^4}{4} + C

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

\therefore f(x) = 8(\log x)^2 - 4(\log x) + 1

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