Question

HELP NEEDED PLS.

Answer 1

\[\int\limits \frac{ t^5}{ \sqrt{t^2+7} } dt\]

Answer 2

what will be the first step ?

Answer 3

well this is what i did...

\[u = t^2 , du = 2tdt\]
\[\frac{ 1 }{ 2 } \int\limits \frac{ u^2 }{ \sqrt{u+7} }\]

and now i am stuck again.

Answer 4

ok, but than you rationalize the denominator not will be more usefully like a first step ?

Answer 5

rationalize the denominator? How so?

Answer 6

@TheSmartOne

Answer 7

eliminate the radical from denominator

Answer 8

So multiply the numerator and denominator by \[\sqrt{t^2+7}\] ?

Answer 9

yes exactly

Answer 10

hi TSO my friend

Answer 11

TSO how you think please this was one bad idea to rationalize the denominator ?

Answer 12

i think this was one simplified step - or not ?

Answer 13

in the next step rewrite the radical in exponential form

Answer 14

Okay...\[\frac{ t^5(t^2+7)^{1/2} }{ t^2+7 }\]

Answer 15

and then what? :/

Answer 16

TSO any idea please ? i ve learned this past 30 years

Answer 17

ago

Answer 18

@agent0smith

Answer 19

Try \( \large u = t^2 + 7 \)

Answer 20

I tired that too, but then it got messy. I got...\[u = t^2+7 \rightarrow t = \sqrt{u-7} \rightarrow du=2tdt\]

\[\frac{ 1 }{ 2 } \int\limits \frac{ \sqrt{u-7}^4 }{ \sqrt{u} }\]

Answer 21

Simplify and expand...

Answer 22

You can also do a trig sub.

Answer 23

It's not hard from here\[\large \frac{ 1 }{ 2 } \int\limits\limits \frac{(u-7)^2 }{ \sqrt{u} } du= \]

Answer 24

put \[\sqrt{t^2+7}=x\]
\[t^2=x^2-7,2t~dt=2x~dx,tdt=xdx\]
\[I=\int\limits \frac{ t^5~dt }{ \sqrt{t^2+7} }=\int\limits \frac{ \left( t^2 \right)^2 tdt }{ \sqrt{t^2+7} }=\int\limits \frac{ \left( x^2-7 \right)^2 }{ x } x~dx\]
can you complete further?

Answer 25

\[= \int\limits \left( x^4-14x^2+49 \right)dx\]

Answer 26

I am trying it @agent0smith 's way.
But would you then factor out the numerator, and then turn the denominator into a negative exponent, then distribute the denominator?

Answer 27

Simplify what i gave as far as possible

Answer 28

okay, so i got\[\frac{ 1 }{ 2 } \int\limits u^{5/2}-14u^{3/2}+49u^{1/2}\]

Answer 29

Are you sure? \[\large \frac{ 1 }{ 2 } \int\limits\limits\limits \frac{(u-7)^2 }{ \sqrt{u} } du= \frac{ 1 }{ 2 } \int\limits\limits\limits \frac{u^2-14u +49 }{ \sqrt{u} } du=\]
\[\large \frac{ 1 }{ 2 } \int\limits\limits \frac{u^2 }{ \sqrt{u} }- \frac{ 14 u }{ \sqrt u } +\frac{ 49 }{\sqrt u } du=\]

Answer 30

Oh oops. forgot to make the exponent negative.

Answer 31

Kay, so i got... \[\frac{ 1 }{ 2 }\left[ \frac{ 2 }{ 5 } u^{5/2}+\frac{ 28 }{ 3 }u^{3/2}+98u^{1/2}\right]\]

Answer 32

after integrating

Answer 33

I got the answer! Thank you very much!