{"id":32281,"date":"2025-06-22T12:18:32","date_gmt":"2025-06-22T12:18:32","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=32281"},"modified":"2025-06-22T12:18:34","modified_gmt":"2025-06-22T12:18:34","slug":"evaluate-the-integral","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/evaluate-the-integral\/","title":{"rendered":"Evaluate the integral"},"content":{"rendered":"\n
\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

We are given the integral:\u222b(2x\u22121)ln\u2061(3x)\u2009dx\\int (2x – 1)\\ln(3x)\\, dx\u222b(2x\u22121)ln(3x)dx<\/p>\n\n\n\n

This integral involves a product of functions, so we apply integration by parts<\/strong>:<\/p>\n\n\n\n

Step 1: Identify parts<\/h3>\n\n\n\n

Let:<\/p>\n\n\n\n

    \n
  • u=ln\u2061(3x)u = \\ln(3x)u=ln(3x) so that du=1xdxdu = \\frac{1}{x} dxdu=x1\u200bdx<\/li>\n\n\n\n
  • dv=(2x\u22121)dxdv = (2x – 1) dxdv=(2x\u22121)dx so that v=x2\u2212xv = x^2 – xv=x2\u2212x<\/li>\n<\/ul>\n\n\n\n

    Step 2: Apply the integration by parts formula<\/h3>\n\n\n\n

    \u222bu\u2009dv=uv\u2212\u222bv\u2009du\\int u\\, dv = uv – \\int v\\, du\u222budv=uv\u2212\u222bvdu<\/p>\n\n\n\n

    Substitute in:\u222b(2x\u22121)ln\u2061(3x)\u2009dx=(x2\u2212x)ln\u2061(3x)\u2212\u222b(x2\u2212x)\u22c51xdx\\int (2x – 1)\\ln(3x)\\, dx = (x^2 – x)\\ln(3x) – \\int (x^2 – x) \\cdot \\frac{1}{x} dx\u222b(2x\u22121)ln(3x)dx=(x2\u2212x)ln(3x)\u2212\u222b(x2\u2212x)\u22c5x1\u200bdx<\/p>\n\n\n\n

    Simplify the second integral:\u222b(x2\u2212x)\u22c51xdx=\u222b(x\u22121)dx=x22\u2212x\\int (x^2 – x) \\cdot \\frac{1}{x} dx = \\int (x – 1) dx = \\frac{x^2}{2} – x\u222b(x2\u2212x)\u22c5x1\u200bdx=\u222b(x\u22121)dx=2×2\u200b\u2212x<\/p>\n\n\n\n

    Step 3: Combine results<\/h3>\n\n\n\n

    \u222b(2x\u22121)ln\u2061(3x)\u2009dx=(x2\u2212x)ln\u2061(3x)\u2212(x22\u2212x)+C\\int (2x – 1)\\ln(3x)\\, dx = (x^2 – x)\\ln(3x) – \\left( \\frac{x^2}{2} – x \\right) + C\u222b(2x\u22121)ln(3x)dx=(x2\u2212x)ln(3x)\u2212(2×2\u200b\u2212x)+C=(x2\u2212x)ln\u2061(3x)\u2212x22+x+C= (x^2 – x)\\ln(3x) – \\frac{x^2}{2} + x + C=(x2\u2212x)ln(3x)\u22122×2\u200b+x+C<\/p>\n\n\n\n

    Final Answer:<\/h3>\n\n\n\n

    (x2\u2212x)ln\u20613x\u2212x22+x+C\\boxed{(x^2 – x)\\ln 3x – \\frac{x^2}{2} + x + C}(x2\u2212x)ln3x\u22122×2\u200b+x+C\u200b<\/p>\n\n\n\n

    Correct multiple choice:<\/h3>\n\n\n\n

    This matches the first choice<\/strong> in the image.Answer: First option\\boxed{\\text{Answer: } \\text{First option}}Answer: First option\u200b<\/p>\n\n\n\n

    Explanation <\/h3>\n\n\n\n

    To evaluate the integral \u222b(2x\u22121)ln\u2061(3x)\u2009dx\\int (2x – 1)\\ln(3x)\\, dx\u222b(2x\u22121)ln(3x)dx, we apply the technique of integration by parts. Integration by parts is used when the integrand is a product of two functions, and the formula is:\u222bu\u2009dv=uv\u2212\u222bv\u2009du\\int u\\, dv = uv – \\int v\\, du\u222budv=uv\u2212\u222bvdu<\/p>\n\n\n\n

    We assign u=ln\u2061(3x)u = \\ln(3x)u=ln(3x), which simplifies upon differentiation. Its derivative is du=1xdxdu = \\frac{1}{x} dxdu=x1\u200bdx. The remaining part, dv=(2x\u22121)dxdv = (2x – 1)dxdv=(2x\u22121)dx, integrates to v=x2\u2212xv = x^2 – xv=x2\u2212x since the integral of 2x2x2x is x2x^2×2, and the integral of \u22121-1\u22121 is \u2212x-x\u2212x.<\/p>\n\n\n\n

    Using the integration by parts formula, we compute uv=(x2\u2212x)ln\u2061(3x)uv = (x^2 – x)\\ln(3x)uv=(x2\u2212x)ln(3x), and then subtract the integral of v\u22c5duv \\cdot duv\u22c5du, which becomes \u222b(x2\u2212x)\u22c51xdx=\u222b(x\u22121)dx\\int (x^2 – x)\\cdot \\frac{1}{x} dx = \\int (x – 1) dx\u222b(x2\u2212x)\u22c5x1\u200bdx=\u222b(x\u22121)dx. This simplifies to x22\u2212x\\frac{x^2}{2} – x2x2\u200b\u2212x. Finally, putting everything together, we get:(x2\u2212x)ln\u2061(3x)\u2212(x22\u2212x)+C(x^2 – x)\\ln(3x) – \\left( \\frac{x^2}{2} – x \\right) + C(x2\u2212x)ln(3x)\u2212(2×2\u200b\u2212x)+C<\/p>\n\n\n\n

    This simplifies to the final answer:(x2\u2212x)ln\u2061(3x)\u2212x22+x+C(x^2 – x)\\ln(3x) – \\frac{x^2}{2} + x + C(x2\u2212x)ln(3x)\u22122×2\u200b+x+C<\/p>\n\n\n\n

    Among the provided choices, this exactly matches the first<\/strong> option in the list.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    The Correct Answer and Explanation is: We are given the integral:\u222b(2x\u22121)ln\u2061(3x)\u2009dx\\int (2x – 1)\\ln(3x)\\, dx\u222b(2x\u22121)ln(3x)dx This integral involves a product of functions, so we apply integration by parts: Step 1: Identify parts Let: Step 2: Apply the integration by parts formula \u222bu\u2009dv=uv\u2212\u222bv\u2009du\\int u\\, dv = uv – \\int v\\, du\u222budv=uv\u2212\u222bvdu Substitute in:\u222b(2x\u22121)ln\u2061(3x)\u2009dx=(x2\u2212x)ln\u2061(3x)\u2212\u222b(x2\u2212x)\u22c51xdx\\int (2x – 1)\\ln(3x)\\, […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-32281","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32281","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=32281"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32281\/revisions"}],"predecessor-version":[{"id":32284,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32281\/revisions\/32284"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=32281"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=32281"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=32281"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}