The discussion centers on evaluating the indefinite integral of the function (2t6 - 3)/t3 dt. A user initially attempted substitution methods with u = t3 and u = 2t6 - 3 but struggled to find appropriate substitutions. The solution provided clarifies that substitution is unnecessary; instead, simplifying the expression to 2t3 - 3t-3 allows for straightforward application of the power rule to find the antiderivatives.
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futurept said:Homework Statement
Evaluate the following indefinite integral: (2t6-3)/t3 dt
The Attempt at a Solution
I know I need to substitute. Tried u= t3 and found du= 3t2dt. Tried to find where du would substitute in, but found nothing. Also tried u= 2t6-3. Found du=12t5dt and again cannot find where to sub in the du. Any suggestions on where I'm going wrong?