Question

Plz solve this and explain all the steps clearly
$integration( \frac{x {}^{2} + 4 }{ {x}^{2} } )$
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Answer 1

Answer :

(x + 1/x) is the required value

Given :

• (x²+4)/x²

To Find :

$\sf \bullet \: \: \int \dfrac{x^{2} + 4}{x^{2}}dx$

Identity to be used :

$\sf \bullet \: \: \int x^{n} = \dfrac{x^{n+1}}{n+1} + C$

Where C be any constant

Solution :

$\sf \longrightarrow \int \dfrac{x^{2}+4}{x^{2}}dx \\\\\sf\longrightarrow \int \{ \dfrac{x^{2}}{x^{2}} + \dfrac{4}{x^{2}} \}\\\\ \sf \longrightarrow\int .1. dx + 4\int x^{-2} dx \\\\ \sf \longrightarrow x + 4\times \{\dfrac{x^{-2+1}}{-2+1} \} + C \\\\ \sf \longrightarrow x - 4x^{-1} + C \\\\ \sf \longrightarrow x - \dfrac{4}{x} + C$

Some other Identities :

$\sf \bullet \: \: \int \dfrac{1}{x}dx = \log_{e} x + C \\\\ \sf \bullet \: \: \int \cos x dx = \sin x + C \\\\ \sf \bullet \: \: \int \sin x dx = -\cos x + C \\\\ \sf \bullet \: \: \int \sec^{2} x dx = \tan x + C \\\\ \sf \bullet \: \: \int \csc^{2} x dx = -\cot x + C \\\\ \sf \bullet \: \: \int \sec x \tan x dx = \sec x + C \\\\ \sf \bullet \: \: \int \csc x \cot x dx = -\cosec x + C \\\\ \sf \bullet \: \: \int e^{x}dx = e^{x} + C \\\\ \sf \bullet \: \: \int a^{x} dx = \dfrac{a^{x}}{log_{e}^{a} } + C$

Answer 2

AnswEr :

Given Integrand,

$\displaystyle \sf \: l = \int \dfrac{ {x}^{2} + 4 }{ {x}^{2} } dx$

Re writting the expression,

$\longrightarrow \: \displaystyle \sf \: l = \int ({x}^{2} + 4) {x}^{ - 2} dx \\ \\ \displaystyle \: \longrightarrow \: \sf \: l \: = \int(1 + {4x}^{ - 2} )dx$

We know that,

• $\displaystyle \sf \int \: x {}^{n} dx = \dfrac{x {}^{n + 1} }{n +1 } + c$

Now,

$\longrightarrow \: \displaystyle \: \sf \: l = \int \: dx + 4 \int \: {x}^{ - 2} dx \\ \\ \longrightarrow \: \displaystyle \: \sf \: l = x + 4 \bigg( - \dfrac{1}{x} \bigg) + c \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf l = x - \dfrac{4}{x} + c }}$