Question

Someone help me with Calculus, please

Answer 1

I might be able to. What is your question?

Answer 2

\[\int\limits \frac{x ^{3}+\sqrt{x} }{ x^{2} }\]

Answer 3

Ok.
\[\int\limits_{}^{}\frac{ x^3+\sqrt{x} }{ x^2 }dx\]
\[\int\limits_{}^{}x^{-2}(x^3+x^{1/2})\]
\[\int\limits_{}^{}x+x ^{-3/2}\]
\[\frac{ x^2 }{ 2 }+\frac{ x ^{-3/2+1} }{ -3/2+1 }+c\]
\[\frac{ x^2 }{ 2 }-\frac{ 2 }{ 3\sqrt{x} }+c\]
I hope that helps and that you were able to follow.

Answer 4

my answer sheet says \[\frac{ 1 }{ 2 }x^{2}-2x^{-1/2}+C\]

Answer 5

Whoops, yeah. I forgot to add the 1 on the bottom of the final expression. That is correct.

Answer 6

Thank you so much<3

Answer 7

\[\frac{ x^2 }{ 2 }-\frac{ 2 }{ \sqrt{x} }\]

Answer 8

No problem.

Answer 9

Could you help me with another problem?

Answer 10

Yes

Answer 11

\[\frac{ d }{ dx } [\ln |\tan x|]\]

Answer 12

:( I accidentally X out of the tab... I have to restart...

Answer 13

Recall that d/dx |u|=u/|u|*u' and d/dx ln(u)=1/u*u'.
\[\frac{ d }{ dx }[\ln |\tan x|]\]
\[\frac{ 1 }{ |\tan x|}*\frac{ d }{ dx }(|\tan x|)\]
\[\frac{ 1 }{ |\tan x|}*\frac{ \tan x }{ |\tan x| }*\frac{ d }{ dx }(\tan x)\]
\[\frac{ 1 }{ |\tan x|}*\frac{ \tan x }{ |\tan x| }*\sec^2x\]
\[\frac{ \tan x*\sec^2x }{ 2|\tan x| }\]

Answer 14

answer sheet says cot x sec^2 x

Answer 15

I realized my mistake. |tanx|*|tanx| doesn't equal 2|tanx|, it would equal |tanx|^2, which is just tan^2x. So,
\[\frac{ \tan x *\sec^2x }{ \tan^2x}\]
\[\frac{ \sec ^2x }{ \tan x }\]
\[\cot x*\sec ^2x\]

Answer 16

Thank you :)

Answer 17

No problem.