Question

$\int (x^3 y^3) \, dx =$
Identify the correct option from the following:

• A. $x^5 \left[ \frac{1}{5} (\log x)^3 - \frac{3}{25} (\log x)^2 + \frac{6}{125} \log x - \frac{6}{625} \right] + c$
• B. $x^5 \left[ \frac{1}{5} (\log x)^3 - \frac{2}{25} (\log x)^2 + \frac{6}{125} \log x - \frac{12}{25} \right] + c$
• C. $x^5 \left[ \frac{1}{5} (\log x)^3 - \frac{4}{25} (\log x)^2 - \frac{6}{125} \log x - \frac{8}{125} \right] + c$
• D. $x^5 \left[ \frac{1}{5} (\log x)^3 + \frac{3}{25} (\log x)^2 - \frac{6}{125} \log x - \frac{6}{25} \right] + c$

Hint:

For integrals of the form $\int x^n (\log x)^m \, dx$, use substitution $u = \log x$ and apply integration by parts repeatedly.

Answer 1

The Correct Option is A

Step 1: Interpret the integrand
The integral $\int (x^3 y^3) \, dx$ likely contains a typo, as $y$ is undefined. Assume $y = \log x$, so the integral is $\int x^3 (\log x)^3 \, dx$, which matches the form of the options. Step 2: Integrate using substitution
Let $u = \log x$, so $x = e^u$, $dx = e^u \, du$. Then $x^3 = e^{3u}$, $(\log x)^3 = u^3$, and the integral becomes $\int e^{3u} u^3 e^u \, du = \int u^3 e^{4u} \, du$. Use integration by parts: $v = u^3$, $dw = e^{4u} \, du$, $dv = 3u^2 \, du$, $w = \frac{1}{4} e^{4u}$. Then $\int u^3 e^{4u} \, du = u^3 \cdot \frac{1}{4} e^{4u} - \int \frac{1}{4} e^{4u} \cdot 3u^2 \, du$. Continue: $\int 3u^2 e^{4u} \, du = 3 \left( u^2 \cdot \frac{1}{4} e^{4u} - \int \frac{1}{4} e^{4u} \cdot 2u \, du \right)$, and so on. Step 3: Compute and match
After three integrations by parts, the result is $e^{4u} \left( \frac{1}{4} u^3 - \frac{3}{16} u^2 + \frac{6}{64} u - \frac{6}{256} \right) + c$. Substitute back: $x^4 \left( \frac{1}{4} (\log x)^3 - \frac{3}{16} (\log x)^2 + \frac{6}{64} \log x - \frac{6}{256} \right) = x^5 \left( \frac{1}{5} (\log x)^3 - \frac{3}{25} (\log x)^2 + \frac{6}{125} \log x - \frac{6}{625} \right) + c$, matching option (1).