1. anti-derivative

i tried to solve this problem using substitution (as per someones suggestion) but the answer my prof came up with differs from mine.

he got $\frac{4}{15}(x^3+3x+2)^\frac{5}{4} + C$

here's the problem:

find the antiderivative

$f(x) = (x^2+1)\sqrt[4]{x^3+3x+2}$

i used substitution

$u = x^3 + 3x + 2$
$h(u) = 3x^2 + 3$

then did:
$
\frac{1}{3}*3(x^2+1)\sqrt[4]{x^3+3x+2}$

which gave me:
$
\frac{1}{3}\sqrt[4]{u} du$

and then:
$
\frac{1}{3}\sqrt[4]{x^3+3x+2} + c$

what am i doing wrong, and how do i get the answer he got?

2. Originally Posted by paguy30
i tried to solve this problem using substitution (as per someones suggestion) but the answer my prof came up with differs from mine.

he got $\frac{4}{15}(x^3+3x+2)^\frac{5}{4} + C$

here's the problem:

find the antiderivative

$f(x) = (x^2+1)\sqrt[4]{x^3+3x+2}$

i used substitution

$u = x^3 + 3x + 2$
$h(u) = 3x^2 + 3$

then did:
$
\frac{1}{3}*3(x^2+1)\sqrt[4]{x^3+3x+2}$

which gave me:
$
\frac{1}{3}\sqrt[4]{u} du$

and then:
$
\frac{1}{3}\sqrt[4]{x^3+3x+2} + c$

what am i doing wrong, and how do i get the answer he got?
let $u = x^3 + 3x + 2$

$du = (3x^{2}+3) dx = 3(x^{2}+1) dx \implies dx = \frac{du}{3(x^{2}+1)}$

$\int (x^{2}+1) \times u^{\frac{1}{4}} \frac{du}{3(x^2+1)} = \frac{1}{3} \int u^{\frac{1}{4}} du$