Now that we know that integration is the reverse of differentiation, we may at once look up the differential coefficients we already know, and see from what functions they were derived. This gives us the following integrals ready made:
x
−
1
(p. 148)
;
∫
x
−
1
d
x
=
log
ϵ
x
+
C
.
1
x
+
a
(p. 149)
;
∫
1
x
+
a
d
x
=
log
ϵ
(
x
+
a
)
+
C
.
ϵ
x
(p. 143)
;
∫
ϵ
x
d
x
=
ϵ
x
+
C
.
ϵ
−
x
∫
ϵ
−
x
d
x
=
−
ϵ
−
x
+
C
{\displaystyle {\begin{alignedat}{4}&x^{-1}\qquad &&{\text{(p. 148)}};&&\qquad &&\int x^{-1}\,dx&&=\log _{\epsilon }x+C.\\&{\frac {1}{x+a}}\qquad &&{\text{(p. 149)}};&&&&\int {\frac {1}{x+a}}\,dx&&=\log _{\epsilon }(x+a)+C.\\&\epsilon ^{x}\qquad &&{\text{(p. 143)}};&&&&\int \epsilon ^{x}\,dx&&=\epsilon ^{x}+C.\\&\epsilon ^{-x}&&&&\int \epsilon ^{-x}\,dx&&=-\epsilon ^{-x}+C\\\end{alignedat}}}
(for it
y
=
−
1
ϵ
x
=
d
y
d
x
=
−
ϵ
x
×
0
−
1
×
ϵ
x
ϵ
2
x
=
ϵ
−
x
).
{\displaystyle {\text{(for it }}y=-{\dfrac {1}{\epsilon ^{x}}}={\dfrac {dy}{dx}}=-{\dfrac {\epsilon ^{x}\times 0-1\times \epsilon ^{x}}{\epsilon ^{2x}}}=\epsilon ^{-x}{\text{).}}}
sin
x
(p. 168)
;
∫
sin
x
d
x
=
−
cos
x
+
C
.
cos
x
(p. 166)
;
∫
cos
x
d
x
=
sin
x
+
C
.
{\displaystyle {\begin{alignedat}{4}&\sin x\qquad &&{\text{(p. 168)}};&&\qquad &&\qquad \int \sin x\,dx&&=-\cos x+C.\\&\cos x\qquad &&{\text{(p. 166)}};&&\qquad &&\qquad \int \cos x\,dx&&=\sin x+C.\\\end{alignedat}}}
Also we may deduce the following:
log
ϵ
x
;
∫
log
ϵ
x
d
x
=
x
(
log
ϵ
x
−
1
)
+
C
{\displaystyle {\begin{alignedat}{4}&\log _{\epsilon }x;&&\qquad &&\qquad \int \log _{\epsilon }x\,dx&&=x(\log _{\epsilon }x-1)+C\\\end{alignedat}}}
(for it
y
=
x
log
ϵ
x
−
x
,
d
y
d
x
=
x
x
+
log
ϵ
x
−
1
=
log
ϵ
x
).
{\displaystyle {\text{(for it }}y=x\log _{\epsilon }x-x,{\dfrac {dy}{dx}}={\dfrac {x}{x}}+\log _{\epsilon }x-1=\log _{\epsilon }x{\text{).}}}