Question

intereration of √x+1/√x​

Answer 1

Answer:-

We have to find:

$\displaystyle \int \sf \bigg(\sqrt{x} + \dfrac{1}{ \sqrt{x} }\bigg) \: \: x$

using ∫ (u ± v) dx = ∫ u dx + ∫ v dx we get,

$\implies \displaystyle \int \sf \sqrt{x} \: \: dx + \displaystyle \int \sf \frac{1}{ \sqrt{x} } \: \: dx$

using n√x = x^(1/n) & 1/a = a⁻¹ we get,

$\implies \displaystyle \int \sf \: {x}^{ \frac{1}{2} } \: \: dx + \displaystyle \int \sf \: {x}^{ \frac{ - 1}{2} } \: \: dx$

using ∫ xⁿ dx = (xⁿ⁺¹) / n + 1 we get,

$\implies \sf \: \frac{ {x}^{ \frac{1}{2} + 1 } }{ \frac{1}{2} + 1 } + \frac{ {x}^{ \frac{ - 1}{2} + 1} }{ \frac{ - 1}{2} + 1 } + c \\ \\ \\ \implies \sf \: \frac{ {x}^{ \frac{1 + 2}{2} } }{ \frac{1 + 2}{2} } + \frac{ {x}^{ \frac{1}{2} } }{ \frac{1}{2} } + c \\ \\ \\ \implies \sf \: {x}^{ \frac{3}{2} } \times \frac{2}{3} + {x}^{ \frac{1}{2} } \times 2 + c \: \: \: \: \: \: ( \because {x}^{ \frac{a}{b}} = \sqrt[b]{ {x}^{a} } \: ) \\ \\ \\ \implies \underline{ \underline{\sf \: \frac{2\sqrt{ {x}^{3} } }{3} + 2 \sqrt{x} + c}}$

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Integrals of some functions:-

• ∫ sin x dx = - cos x + c

• ∫ cos x dx = sin x + c

• ∫ tan x dx = log (sec x) + c

• ∫ a dx = ax + c (where a is constant)

• ∫ sin ax = (- cos ax)/a + c

• ∫ cos ax = (sin ax)/a + c