Question

integral (sq rt of x to the 4th root -6x^2+e^x-2/x) dx

Answer 1

can you use draw box below to draw it out? I don't understand your problem

Answer 2

\[\int\limits \left( \sqrt[4]{x}-6x^2 +e^x-2/x \right)\]

Answer 3

I was looking for that to type my equation when i posted my question originally and couldnt find it

Answer 4

\[\int\limits \left( x^{\frac{1}{4}}-6x^2 +e^x-2x^{-1}\right)dx\]
yeah, easy now, right?

Answer 5

thats about as far as i got on the problem

Answer 6

wasn't sure if i take the antiderivative next or not

Answer 7

\[=\frac{x^{\frac{1}{4}+1}}{\frac{1}{4}+1}-4x^3+e^x-2lnx+C\]

Answer 8

simplify the first term
\[\frac{4}{5}*x^{\frac{5}{4}}-4x^3+e^x-2lnx+C\]

Answer 9

that's it

Answer 10

ok, so i see you combined the two x's to get a 4 instead of a 12 like i did

Answer 11

I have class now, cannot explain . sorry for that.

Answer 12

@UnkleRhaukus please take it over,

Answer 13

@ganeshie8 Please

Answer 14

i donno why but i got \[\frac{ 1 }{ 4 }x^{-3/4}-12x+e^x+2x ^{-2}\]

Answer 15

\[\int\limits \left( \sqrt[4]{x}-6x^2 +e^x-\frac2x \right)\,\mathrm dx\\
=\int\limits \left( x^{\frac{1}{4}}-6x^2 +e^x-2x^{-1}\right)\,\mathrm dx\\
=\int\limits x^{\frac{1}{4}}\,\mathrm dx-\int\limits6x^2\,\mathrm dx+\int\limits e^x\,\mathrm dx-2\int\frac {\mathrm dx}x\\
=\frac{x^{\frac{1}{4}+1}}{\frac{1}{4}+1}-\frac{6x^{2+1}}{2+1}+e^x-2\ln x+C\\=\]

Answer 16

you are differentiation where you should be integrating

Answer 17

OH I see! Thank you

Answer 18

\[\frac{\mathrm d}{\mathrm dx}(x^n+C)=nx^{n-1}\\----------\\
\int x^n\mathrm dx=\frac{x^{n+1}}{n+1}+C\]

Answer 19

ok thank you!

Answer 20

you notice that if the index \(n\) is -1,


\[\begin{align}\int x^{-1}\mathrm dx&=\color{red}{\cancel{\color{black}{\frac{x^{-1+1}}{-1+1}+C =\frac{1}{0}+C???}}}\\
\\&=\ln x+C\\
\end{align}\]

it dosent work, you have to

Answer 21

... know this .