How do you integrate $intx(4x+5)^3$ using substitution?

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Answer 1 · 2017-02-07T18:05:16

$=>1/80(4x + 5)^5 - 5/64(4x + 5)^4 + C$

Let $u = 4x + 5$ . Then $du =4dx -> dx= (du)/4$ . This also means that $x = (u - 5)/4$ .

$=>int (u - 5)/4 u^3 (du)/4$

$=>1/16intu^4 - 5u^3du$

$=>1/16(1/5u^5 - 5/4u^4) + C$

$=>1/80u^5 - 5/64u^4 + C$

$=>1/80(4x + 5)^5 - 5/64(4x + 5)^4 + C$

Hopefully this helps!

How do you integrate intx(4x+5)^3intx(4x+5)^3 using substitution?