### hwk6, solutions - Tomasz Przebinda

```MATH. 2924, HOMEWORK 6, SOLUTIONS DUE FRIDAY 9.26.2014
TOMASZ PRZEBINDA
1. Which of the following rational functions are simple fractions
2
2
2x
2
, (b)
, (c)
,
(a)
, (d) 2
2
x−2
x−2
(x − 2)
x − 3x + 2
(e)
(x2
2x + 2
.
− x + 2)3
(b), (c), (e).
2. Decompose into partial fractions over R:
2x2 + 2x + 13
.
x5 − 2x4 + 2x3 − 4x2 + x − 2
1
x+2
3x + 4
− 2
− 2
.
x − 2 x + 1 (x + 1)2
3. Compute the integral
Z
ln(x2 − x + 2) dx.
We do it in stages. First integration by parts gives
Z
Z
2x − 1
2
2
ln(x − x + 2) dx = xln(x − x + 2) − x 2
dx.
x −x+2
Next, by long division
2x − 1
2x2 − x
x−4
=
=2+ 2
.
2
2
x −x+2
x −x+2
x −x+2
Here the last fraction is simple and
1
7
x2 − x + 2 = (x − )2 + .
2
4
Therefore we transform this simple fraction as follows
x
(x − 12 ) − 72
x − 12
x−4
=
=
x2 − x + 2
(x − 12 )2 + 47
(x − 12 )2 +
1
7
4
−
(x −
(1)
(2)
7
2
1 2
)
2
+
7
4
2
TOMASZ PRZEBINDA
Furthermore
Z
and
Z
(x −
7
2
1 2
)
2
+
x − 12
(x − 12 )2 +
7
4
7
dx =
2
r
7
4
7
1
dx = ln((x − )2 + )
2
4
√
x− 1
4
2x − 1
tan−1 ( q 2 ) = 7 tan−1 ( √ )
7
7
7
4
Hence the integral of (2) is equal to
√
7
2x − 1
1
2x + ln((x − )2 + ) − 7 tan−1 ( √ ).
2
4
7
Therefore (1) is equal to
√
1
7
2x − 1
xln(x2 − x + 2) − 2x − ln((x − )2 + ) + 7 tan−1 ( √ ).
2
4
7
4. Compute the integral
Z
x tan−1 (x) dx
Again integration by parts show that
Z
Z
1 2
1 2 1
−1
−1
x tan (x) dx = x tan (x) −
x
dx
2
2 x2 + 1
Furthermore
1
x2
x2 + 1 − 1
1
x2 2
= 2
=
=1− 2
.
2
x +1
x +1
x +1
x +1
Thus
Z
Z
1 2 1
1 1 1
x 1
x 2
dx = ( −
) dx = − tan−1 (x).
2
2 x +1
2 2x +1
2 2
Finally
Z
x 1
1
x tan−1 (x) dx = x2 tan−1 (x) − + tan−1 (x).
2
2 2
6. Compute the integral
Z
4x2 + 4x − 11
dx.
(2x − 1)(2x + 3)(2x − 5)
Here is the partial fraction decomposition of the function under the integral
4x2 + 4x − 11
1 1
1 1
3 1
=
.
1 −
3 +
(2x − 1)(2x + 3)(2x − 5)
4x− 2
8x+ 2
8 x − 25
MATH. 2924, HOMEWORK 6, SOLUTIONS DUE FRIDAY 9.26.2014
Hence
Z
3
4x2 + 4x − 11
1
1
1
3
3
5
dx = ln(|x − |) − ln(|x + |) + ln(|x − |) + C.
(2x − 1)(2x + 3)(2x − 5)
4
2
8
2
8
2
7. Compute the integral
Z
1
dx.
x4 + 1
Notice that
x4 + 1 = (x4 + 2x2 + 1) − 2x2 = (x2 + 1)2 − 2x2 = (x2 −
√
2x + 1)(x2 +
Hence we get the partial fraction decomposition
√
√
x+ 2
x− 2
1
1
1
√
√
= √
− √
x4 + 1
2 2 x2 + 2x + 1 2 2 x2 − 2x + 1
Therefore the integral we are computing is equal to
√
√
1
x2 + 2x + 1
1
2x
√ ln(
√
) + √ tan−1 (
) + C.
2
1 − x2
4 2 x − 2x + 1
2 2
√
2x + 1).
```