Explanation Formula used: Integration by parts: “Suppose u and v are differentiable functions, then ∫ u d v = u v − ∫ v d u ”. Calculation: Given that u = ln x and d v = x d x . Integrate the term d v = x d x to obtain the value of v . v = ∫ x d x = x 2 2 Thus, the value of v = x 2 2 . The derivative of u = ln x is d u = 1 x d x . Use the integration by parts ∫ u d v = u v − ∫ v d u to compute the value of ∫ x ln x d x

Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

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Chapter 8.2, Problem 3E

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To evaluate: The value of ∫xlnxdx with u=lnx and dv=xdx by using integration by parts.

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Chapter 8.2, Problem 2E

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Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

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The value of ∫ x ln x d x with u = ln x and d v = x d x by using integration by parts.

Solution Summary: The author evaluates the value of displaystyle Xmathrmlnxdx by using integration by parts.

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To evaluate: The value of ∫xlnxdx with u=lnx and dv=xdx by using integration by parts.

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