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Answer Key for Exam 1
1(i) If we let w = x^3 then dw = 3x^2 dx, so dw 3 = x^2 dx and we have ∫ x^2 cos
x^3
dx =
cos w dw =
sin w + C =
sin
x^3
+ C.
1(ii) If we let w = x^2 then dw = 2x dx, so dw 2 = x dx and we have ∫ x^3 cos
x^2
dx =
x^2 cos
x^2
x dx =
w cos w dw.
Integrate this by parts with dv = cos w dw and u = w, so that du = dw and v = sin w and we have ∫ x^3 cos
x^2
dx =
w cos w dw
=
w sin w −
sin w dw
=^1
(w sin w − − cos w) + C
=^1 2 (w sin w + cos w) + C
=^1 2
[
x^2 sin
x^2
x^2
)]
+ C.
2(a) Integrate by parts with dv = ex^ dx and u = x, so that v = ex^ and du = dx and we have ∫ xex^ dx = xex^ −
ex^ dx = xex^ − ex^ + C = (x − 1)ex^ + C.
2(b) By the product rule we have
d dx xex^ = x (ex)′^ + (x)′^ ex^ = xex^ + 1 · ex^ = (x + 1)ex.
This is no surprise given the answer to (a).
2(c) If we integrate by parts with u = xex^ and dv = dx (x + 1)^2 = (x + 1)−^2 dx, then v = −(x + 1)−^1 =
x + 1 and, by 2(b), du = (x + 1)ex^ dx. Then we have ∫ xex (x + 1)^2 dx = xex
x + 1
∫ (^
x + 1
(x + 1)ex^ dx = − xe
x x + 1
ex^ dx = − xe
x x + 1
Another method is to try the same trick as on the integral of the week #2. We can write ∫ (^) xex (x + 1)^2 dx =
∫ (^) (x + 1) − 1 (x + 1)^2 ex^ dx =
∫ (^) ex x + 1 dx −
∫ (^) ex (x + 1)^2 dx.
If we now integrate the first term on the right by parts with u = 1 x + 1 and dv = ex^ dx then v = ex^ and
du =
(x + 1)^2 and we get
∫ xex (x + 1)^2 dx =
ex x + 1
−ex (x + 1)^2 dx
ex (x + 1)^2 dx.
The last two integrals cancel, so we conclude that ∫ xex (x + 1)^2 dx = e
x x + 1
+ C.
This is the same as the previous answer since
− xe
x x + 1
x x + 1
- ex^ x^ + 1 x + 1 = ex^ x^ + 1^ −^ x x + 1 = e
x x + 1
- We can start by letting u = ln x, so that du = (^) x^1 dx = dxx. If x = e then u = ln e = 1, and if x = ∞ then u = ln ∞ = ∞, so we have (^) ∫ (^) ∞
e
dx x
[
1 + (ln x)^2
] =
1
du 1 + u^2
If you don’t remember what this equals you can look it up (or substitute u = tan θ); it’s an arctangent. So we have (^) ∫ (^) ∞
e
dx x
[
1 + (ln x)^2
] (^) = arctan u|∞ 1 = arctan ∞ − arctan 1 = π 2 −^
π 4 =^
π
Personally, I would have started by letting ln x = tan θ, but that would not be fundamentally different than what we did.
4(i) After we write (^) ∫ ∞ 1
dx 1 + ex^
1
dx 1 + ex
e−x e−x^
1
e−x^ dx e−x^ + 1
we can see that the substitution u = e−x^ + 1 is reasonable. Then du = −e−x^ dx, so that −du = e−x^ dx. If x = 1 then u = e−^1 + 1 = (^1) e + 1, and if x = ∞ then u = e−∞^ + 1 = 0 + 1 = 1, so we get
∫ (^) ∞
1
dx 1 + ex^ =
1
e−x^ dx e−x^ + 1
= −
1+ (^1) e
du u
=
∫ (^) 1+ (^1) e
1
du u = ln u|1+^
(^1) e 1 = ln
e
− ln 1
= ln
e
We could also rewrite the answer as ∫ (^) ∞
1
dx 1 + ex^ = ln
e + 1 e
= ln(e + 1) − ln e = ln(e + 1) − 1.
4(ii) Since 1 1 + ex^ is positive and decreasing for all x we know that
∑^ ∞ n=
1 + en^ and
1
dx 1 + ex^ are either both finite or both infinite.
