Question

\[\int\limits_{}^{} \frac{ x^2 + 3x + 7 }{ \sqrt{x}} dx\]
Using u-sub?

Answer 1

\(\sqrt{x} = u\)

Answer 2

Try using \[
\int x^n\;dx = \frac{x^{n+1}}{n+1}
\]As well as: \[
\int f(x)+g(x)\;dx = \int f(x)\;dx+\int g(x)\;dx
\]

Answer 3

I don't see how i'm supposed to use the sum rule for this one wio o.O
ganeshie i also don't see how i can do this with u = sqrt(x)..

Answer 4

\[
\frac{ x^2 + 3x + 7 }{ \sqrt{x}}
=\frac{x^2}{\sqrt x}+\frac{3x}{\sqrt x}+\frac{7}{\sqrt x}
\]

Answer 5

oh. well this is going to a lot of work..

Answer 6

u = sqrt(x)

du = 1/sqrt(x) dx

Answer 7

u^2 = x

Answer 8

both methods are same work...

Answer 9

* du = 1/2sqrt(x) dx

Answer 10

\[
\frac{a^n}{\sqrt x} = ax^n(x^{-1/2})=ax^{n-1/2}
\]

Answer 11

I'm supposed to use u-sub for this one though.
so u = sqrt(x)
du = 1/sqrt(x) dx
Could you walk me through this with u-sub?

Answer 12

But I don't see any U sub that makes this easier than it already is.

Answer 13

*du = 1/2sqrt(x) dx

Answer 14

\(\large u = \sqrt{x} \)

\(\implies \)
\(\large du = \frac{1}{2\sqrt{x}}dx \)

\(\large x = u^2\)

Answer 15

substitute them in the integral

Answer 16

^^as wio said, both methods involve same effort

Answer 17

\(\int\limits_{}^{} \frac{ x^2 + 3x + 7 }{ \sqrt{x}} dx \)

changes to

\(2\int\limits_{}^{} u^4 + 3u^2 + 7~ du \)