Substitute
and
. Then the integral transforms to

Apply the power rule.

Now put the result back in terms of
.

Answer:
500
Step-by-step explanation:
Answer:
I cant see it
Step-by-step explanation:
Answer:
The answer in simplest term is 35.33333334.
Step-by-step explanation:
First, we need to divide 71 and 6.
71/6 = 23.66666667.
Now we need to divide 35 and 6.
35/6 = 11.66666667.
Now we need to add the two quotients.
23.66666667 + 11.66666667 = 35.33333334.
And that is the final answer.
I hope this helps!


Consequently, t<span>he limit of

as x approaches infinity is

.
In other words,

approaches the line y=x,
</span><span>
so oblique asymptote is y=x.
I'm Japanese, if you find some mistakes in my English, please let me know.</span>