How do you integrate $y = (5x + 3)^6$ ?

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Answer 1 · 2016-11-01T01:55:53

$=1/35*(5x+3)^7+C$

So you've got...

$int(5x+3)^6dx$

Now, make:

$u=5x+3$

Which means that:

$(du)/(dx)=5$

And as a result:

$(dx)/(du)=1/5$

$dx=1/5du$

Now, transform:

$int(5x+3)^6dx$ into $1/5intu^6du$

Integrate:

$1/5intu^6du$

$=1/5*u^7/7+C$

$=1/35*u^7+C$

$=1/35*(5x+3)^7+C$

$=int(5x+3)^6dx$

How do you integrate y = (5x + 3)^6y = (5x + 3)^6?