Question

help finding antiderivative

Answer 1

\[(x ^{3}-3\sqrt{x}+1)/x^{2}\] @mathmate

Answer 2

no answers plz

Answer 3

Recall that \(\sqrt{x}=x^{1/2}\). So it follows that

\(\large \begin{aligned}\dfrac{x^3-3\sqrt{x}+1}{x^2} &= \dfrac{x^3-3x^{1/2}+1}{x^2}\\ &= \dfrac{x^3}{x^2} - 3\dfrac{x^{1/2}}{x^2} + \dfrac{1}{x^2}\\ &= x - 3x^{-3/2} + x^{-2}\end{aligned}\)

What do you get when you integrate the simplified form? :-)

Answer 4

Hint: divide each term and integrate each term separately as a power of x.

Answer 5

\[x ^{2}+6x - \frac{ 1 }{ x }\]

Answer 6

6/x^1/2

Answer 7

\[\int\limits_{}^{}x^3/x^2dx-\int\limits_{}^{}3\sqrt{x}/x^2+\int\limits_{}^{}1/x^2\]
\[\int\limits\limits_{}^{}x=x ^{1+1}/2=x^2/2\]
\[3\int\limits_{}^{}x ^{1/2}/x^2=3\int\limits_{}^{}x ^{-3/2}=3*x ^{-3/2+1/-1/2}=-6/\sqrt{x}\]
\[\int\limits_{}^{}x^-2=x ^{-2+1}/-2+1=-1/x\]

Answer 8

gotta power through antiderivatives before i can answer with just integrals

Answer 9

@robertrico
After all this juggling with integrals, do recall that antiderivatives require an integration constant (+C).

Answer 10

yea that integral only got me to f prime. im doing another of that

Answer 11

all over the place