document.write( "Question 407767: Evaluate the integral using integration by parts with explanations.\r
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Algebra.Com's Answer #287367 by richard1234(7193)\"\" \"About 
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This is an alternative approach which can still use integration by parts, but is slightly easier.\r
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\n" ); document.write( "\n" ); document.write( "Let \"u+=+2x%2B1\", \"du+=+2dx\", \"dx+=+du%2F2\". If we replace 2x+1, we obtain\r
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\n" ); document.write( "\n" ); document.write( "\"%281%2F2%29int%28ln%28u%29%2C+du%29\". We can integrate \"int%28ln%28u%29%2C+du%29\" via parts, or by a formula which you can (and should) memorize.\r
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\n" ); document.write( "\n" ); document.write( "Letting \"y+=+ln+%28u%29\", \"dz+=+du\", we obtain \"dy+=+du%2Fu\" and \"z+=+u\". Applying the parts formula,\r
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\n" ); document.write( "\n" ); document.write( "\"int%28ln%28u%29%2C+du%29+=+u%2Aln+%28u%29+-+int%28u%2Fu%2C+du%29+=+u%2Aln+%28u%29+-+u+%2B+C\". Note that we have to multiply by 1/2, since we want \"%281%2F2%29int%28ln%28u%29%2C+du%29\". Back-substitute \"u+=+2x%2B1\" to obtain\r
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