1
Antiderivatives, indefinite integral
Basic formulas:
Z
Z
0 dx = C
Z
Z
1 dx = x + C
xr+1
x dx =
+ C, pro r 6= −1 a pro ppustn x
r+1
Z
r
x
1
dx = ln |x| + C, kde x 6= 0
x
Z
x
ax dx =
e dx = e + C
Z
Z
sin x dx = − cos x + C
cos x dx = sin x + C
Z
Z
ax
+ C, pro a > 0, a 6= 1
ln a
Z
1
√
dx = arcsin x + C, kde x ∈ (−1, 1)
1 − x2
x2
Z
1
π
dx
=
tg
x
+
C,
kde
x
=
6
+ kπ, k ∈ Z
cos2 x
2
1
dx = arctg x + C
+1
1
dx = −cotg x + C, kde x 6= kπ, k ∈ Z
sin2 x
1)
Z
√
( x + 3x + 2x − 4) dx =
3
=
x2
3
2
Z
1
2
Z
x dx +
Z
x
3 dx + 2
Z
x dx − 4
1 dx =
x2
2 √ 3
3x
3x
+
+2·
− 4x + C = · x +
+ x2 − 4x + C.
ln 3
2
3
ln 3
2)
Z
2
cos2 x
dx =
sin2 x
Z
cotg x dx =
Z
1 − sin2 x
dx =
sin2 x
Z
1
dx −
sin2 x
Z
1 dx = −cotg x − x + C.
3)
Z
(x − 1)2
dx =
x3
Z
x2 − 2x + 1
dx =
x3
= ln |x| − 2 ·
Integration by parts
Z 1
2
1
− 2+ 3
x x
x
Z
dx =
1
dx −2
x
x−1
x−2
2
1
+
+ C = ln |x| + − 2 + C,
−1
−2
x 2x
Z
x 6= 0.
x
−2
Z
dx +
x−3 dx =
2
Z
Z
0
f ·g =f ·g−
f0 · g
R
1) x · ex dx = I
Integration by parts:
f = x,
g 0 = ex .
f 0 = 1,
g = ex .
Z
I=
2)
R
x
x
Z
xe dx = xe −
ex dx = x · ex − ex + C = ex (x − 1) + C.
x2 sin x dx = I.
f = x2
Let
g 0 = sin x,
then
f 0 = 2x
g = − cos x.
R
R
R
I = x2 sin x dx = −x2 cos x + 2 x cos x dx . For x cos x dx use again integration by
parts:
f =x
f0 = 1
g 0 = cos x
g = sin x
Z
Z
x cos x dx = x sin x − sin x dx = x sin x + cos x + C.
Substitute:
3)
R
Z
x2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + C.
ln x dx =,
Z
Z
Z
1
· x dx = x · ln x − 1 dx = x ln x − x + C,
1 · ln x dx = x · ln x −
x
where
f = ln x
g0 = 1
1
x
g = x.
f0 =
3
Exercises
1 Evaluate integrals.
a)
Z 1
2
√ − 5x +
x
x
Z
c)
Z
dx
b)
Z
2
tg x dx
d)
(2 − x)2
dx
x3
√
x3x
√ dx
x
2 Integrate by parts.
Z
x
(1 − x)e dx
a)
Z
c)
Z
e)
ln x
dx
x2
ln x3 dx
Z
b)
Z
d)
Z
f)
xe−x dx
√
3
x ln x dx
x2 cos x dx
Solutions
1
√
5x
a) 2 x −
+ 2 ln |x| + C
ln 5
c) tg x − x + C
4
2
− 2 + ln |x| + C
x
x
6 √
6
d)
x11 + C
11
b)
2
a) (2 − x)ex + C
1
c) (− ln x − 1) + C
x
e) x(ln x3 − 3) + C
b) (−x − 1)e−x + C
3√
3
3
d)
x4 ln x −
+C
4
4
f) x2 sin x + 2x cos x − 2 sin x + C