Question

d/dx * (integral of ln(t^2+1)dt from 0 to x^3) is? I will write it with the quation thing

Answer 1

\[d/dx ( \int\limits_{0}^{x^3} \ln (t^2 +1)dt) \]

Answer 2

don't even think about integrating that

Answer 3

plug in x^3 for t

Answer 4

\[d/dx \int\limits_{0}^{x^3} f(t )dt\]

\[d/dx |_{0}^{x^3} F(t)dt\]

\[d/dx ( F(x^3) - F(0) dt)=d/dx F(x^3)= 3x^2 f(x^3)\]

Answer 5

\[f(t)=ln((t^2+1))\]

f(x^3)=ln((x^6+1))

answer should be (from above post)

\[3x^2 ln (x^6+1)\]

Answer 6

thanks!!! =)

Answer 7

it was really confusing me

Answer 8

i was sittin trying to integrate XD

Answer 9

Generally you see d/dx and integral , you don't have to integrate. Like for this problem, integral for kind of complicated if we integrate

Answer 10

can i ask u a question about integrals?

Answer 11

yes

Answer 12

when u have to integrate two functions that are being multiplied how shud u go about doing it? For example,
\[\int\limits_{0}^{1} (x+1) \sqrt{x} dx \]

Answer 13

for this particular problem you are better off distributing it

\[(x+1)x^{1/2}=x^{3/2}+x^{1/2}\]