Showing all your working, evaluate ∫(21x^6 - e^2x- (1/x) +6)dx

 ∫(21x6-e2x-1/x+6)dx

To tackle this question, I would do each part separately.

Firstly, take 21x6...using the product rule this intergrates like so (21/(6+1))x6+1 = (21/7)x7 = 3x7

Second, ekx always intergrates to (1/k)ekx .........so -e2x goes to (-1/2)e2x.

1/x has the common intergral of lnx which must simply be learnt.

And finally, 6 intergrates to 6x.

We then collect all these parts and put them together so we get y=3x7-(1/2)e2x-lnx+6x+C

Always remember that for intergration without limits you must always add C at the end.

DG
Answered by David G. Maths tutor

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