int (1 - x^2) log x , dx = | vedclass

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(A) We need to evaluate the integral $I = \int (1 - x^2) \log x \, dx$. Using the distributive property,we can write this as $I = \int \log x \, dx -

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$\left( x - \frac{x^3}{3} \right) \log x - \left( x - \frac{x^3}{9} \right) + c$

Vedclass · Answer

(A) We need to evaluate the integral $I = \int (1 - x^2) \log x \, dx$. Using the distributive property,we can write this as $I = \int \log x \, dx - \int x^2 \log x \, dx$. For the first part,$\int \log x \, dx = x \log x - x$. For the second part,we use integration by parts $\int u \, dv = uv - \int v \, du$. Let $u = \log x$ and $dv = x^2 \, dx$. Then $du = \frac{1}{x} \, dx$ and $v = \frac{x^3}{3}$. So,$\int x^2 \log x \, dx = \frac{x^3}{3} \log x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx = \frac{x^3}{3} \log x - \frac{1}{3} \int x^2 \, dx = \frac{x^3}{3} \log x - \frac{x^3}{9}$. Combining these results: $I = (x \log x - x) - \left( \frac{x^3}{3} \log x - \frac{x^3}{9} \right) + c$. Rearranging the terms: $I = \left( x - \frac{x^3}{3} \right) \log x - x + \frac{x^3}{9} + c$. This can be written as $I = \left( x - \frac{x^3}{3} \right) \log x - \left( x - \frac{x^3}{9} \right) + c$.

$\int (1 - x^2) \log x \, dx = $

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