Question

simplify the following surd expressions 6 root 2 - 5 root 2 and root 11 + 3 root 11

Answer 1

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

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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