Question

Integration 2x by (1+xsquare )^1/4 dx

Answer 1

Question :–

$\\ { \bold{Find \: \: the \: \: value : \int \dfrac{2x}{ {(1 + {x}^{2} )}^{ \frac{1}{4} } }.dx }}$

ANSWER :–

$\\ { \bold{I = \int \dfrac{2x}{ {(1 + {x}^{2} )}^{ \frac{1}{4} } }dx }}$

▪︎Now let's use substitution method –

• Let's put $\: \: { \bold{1 + {x}^{2} = t \: \: - }} \: \:$

• Now Differentiate with respect to 't' –

$\\ \implies \: { \bold{(2x) \dfrac{dx}{dt} = 1 \: \: }} \: \:$

$\\ \implies \: { \bold{(2x)dx= dt \: \: }} \: \:$

▪︎ So that –

$\\ \implies { \bold{I = \int \dfrac{dt}{ {(t)}^{ \frac{1}{4} } } }}$

$\\ \implies { \bold{I = \int {(t)}^{ - \frac{1}{4} } dt}}$

$\\ \implies { \bold{I = \ \dfrac{ {t}^{( - \frac{1}{4} + 1)} }{ - \frac{1}{4} + 1} + c}}$

$\\ \implies { \bold{I = \ \dfrac{ {t}^{\frac{3}{4}} }{ \frac{3}{4} } + c}}$

$\\ \implies { \bold{I = \frac{4}{3} { {t}^{\frac{3}{4}} } + c}}$

• Now replace 't' –

$\\ \implies { \red{ \boxed{ \bold{I = \frac{4}{3} { {(1 + {x}^{2} )}^{\frac{3}{4}}} + c} }}}$

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$\\ { \pink{ \bold{ \underline{Used \: \: formula} : - }}}$

$\\ { \bold{(1) \: \: \frac{d( {x}^{n} )}{dx} = n {x}^{n - 1} }}$

$\\ { \bold{(2) \: \: \frac{d(x )}{dx} = 1 }}$

$\\ { \bold{(3) \: \: \int {x}^{n}dx = \frac{ {x}^{n + 1} }{n + 1} + c }}$

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