Question

Integration of x to the power of - 1 dx

Answer 1

$\implies \sf \displaystyle \int \sf{x}^{ - 1} dx$

We know that :-

$\implies \sf \boxed{ \displaystyle \sf{a}^{ - 1} = \frac{1}{a} }$

$\implies \sf \displaystyle \int \sf{x}^{ - 1} dx = \sf \displaystyle \int \sf \dfrac{1}{x} dx$

$\implies\sf \displaystyle \int \sf \dfrac{1}{x} dx = log(x) + c$

Where c is constant.

Answer :

$\sf \displaystyle \int \large\sf{x}^{ - 1} dx = \large \sf \: log(x) + c$

_______________________

Learn more :-

∫ 1 dx = x + C

∫ sin x dx = – cos x + C

∫ cos x dx = sin x + C

∫ sec²x dx = tan x + C

∫ csc²x dx = -cot x + C

∫ sec x (tan x) dx = sec x + C

∫ csc x ( cot x) dx = – csc x + C

∫ (1/x) dx = ln |x| + C

∫ ex dx = ex+ C

∫ ax dx = (ax/ln a) + C

Answer 2

Answer:

⟹∫x

−1

dx

We know that :-

\implies \sf \boxed{ \displaystyle \sf{a}^{ - 1} = \frac{1}{a} }⟹

a

−1

=

a

1

\implies \sf \displaystyle \int \sf{x}^{ - 1} dx = \sf \displaystyle \int \sf \dfrac{1}{x} dx⟹∫x

−1

dx=∫

x

1

dx

\implies\sf \displaystyle \int \sf \dfrac{1}{x} dx = log(x) + c⟹∫

x

1

dx=log(x)+c

Where c is constant.

Answer :

\sf \displaystyle \int \large\sf{x}^{ - 1} dx = \large \sf \: log(x) + c∫x

−1

dx=log(x)+c

_______________________

Learn more :-

∫ 1 dx = x + C

∫ sin x dx = – cos x + C

∫ cos x dx = sin x + C

∫ sec2 dx = tan x + C

∫ csc2 dx = -cot x + C

∫ sec x (tan x) dx = sec x + C

∫ csc x ( cot x) dx = – csc x + C

∫ (1/x) dx = ln |x| + C

∫ ex dx = ex+ C

∫ ax dx = (ax/ln a) + C