Integral of x^3/(1+x^4) (substitution)
Integral of x^3/(1+x^4) (substitution)

## Some Review Problems Exam 2

• Find y’ at x = 1/2, for y = f(sin-1(x)),
where f(x) is a function such that

f ‘(Pi/6) = 7.

By the chain rule,
y’ = f ‘(sin -1(x))d(sin -1(x))/dx
= f ‘(sin -1(x))/(1 – x2)1/2.
At x = 1/2 we get y’ = f'(Pi/6)*2/31/2 = 7*2/31/2.
• Find the exact value of:
tan(sin-1(1/3)) = 1/81/2
• Simplify:
sin(sec-1(x)) = (x2 – 1)1/2/x
• Evaluate:

Use the substitutions u = sin(x), and cos2(x) =
1 – sin2(x) to get
(sin3(x))/3 – (sin5(x))/5 + C.
• Evaluate using integration by parts:

Pick dv = e2xdx, u = x, to get
(x/2)e2x – (1/4)e2x + C
• Evaluate using completing the square:

After completing the square, what we need to integrate
is x/((x+1)2+1) dx. Now substitute u = x + 1 to get
(u – 1)/(u2+1) du. This gives two integrals,
u/(u2+1) du and -1/(u2+1) du.
The first gives (1/2) ln |u2+1| + C,
the second gives -arctan(u) + C. Thus the answer to the original
problem is (1/2) ln |(x+1)2+1| – arctan(x+1) + C.
• Evaluate using partial fractions:

The answer is ln |x| + ln |x+2| + C.
• Determine whether the following improper

• This one converges: using the substitution u = x3 – 1,
the integral is limt -> 1+
(2/3)(x3 – 1)1/2|t2 =
(2/3)71/2 – 0.

This one converges, too:
the integral is limR -> infinity
arctan(x)|0R =
Pi/2 – 0.

• Use a comparison to determine whether the following improper

Since e-u is decreasing, and since
x < x3/2 when x >= 1, we know
e-x > e-x3/2
over the interval of integration,
hence the given integral converges if the integral
of e-x converges. But the integral of e-x
from 1 to infinity does converge; it is equal to
limR -> infinity
-e-x|1R = -0 – -e-1
= 1/e.
• Use a comparison to determine whether the following improper

&nbsp&nbsp (x2 + x)/(x3 + 2) dx
1

Whether the integral converges or not depends on what hapens when x is very large. But for large x, x is negligible compared to x2,
so (x2 + x) is about the same as x2.
Likewise, (x3 + 2) is about the same as
x3. Thus (x2 + x)/(x3 + 2)
is about the same as x2/x3 = 1/x,
and we know the integral of 1/x from 1 to infinity diverges.
But that was just a heuristic argument. Here’s the rigorous
argument using the comparison theorem. Since we now expect
that we want to show divergence, we need to find an f(x) which
is less than (x2 + x)/(x3 + 2) but
for which the integral diverges.
Note that (x2 + x)/(x3 + 2) >
x2/(x3 + 2) >
x2/(x3 + 2×3)
= x2/(3×3)
= (1/3)(1/x). But of course the integral from 1 to infinity
of (1/3)(1/x) diverges, since it does for
1/x (one third of infinity is still infinity!).

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