Question

express the integration of the given function with respect to dx
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Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\displaystyle\int\sf \: \dfrac{cosx \: + \: x \: sinx}{x(x + cosx)} \: dx$

Let assume that

$\rm :\longmapsto\:I \: = \: \displaystyle\int\sf \: \dfrac{cosx \: + \: x \: sinx}{x(x + cosx)} \: dx$

On adding and Subtracting 'x' in numerator, we get

$\rm :\longmapsto\:I \: = \: \displaystyle\int\sf \: \dfrac{cosx \: + \: x \: sinx \: + x \: - \: x}{x(x + cosx)} \: dx$

On rearranging the terms in numerator,

$\rm :\longmapsto\:I \: = \: \displaystyle\int\sf \: \dfrac{(x + cosx) \: + (\: x \: sinx \: - \: x)}{x(x + cosx)} \: dx$

$\rm :\longmapsto\:I \: = \: \displaystyle\int\sf \: \dfrac{(x + cosx) \: - \: x (\: 1 - \: sinx)}{x(x + cosx)} \: dx$

On splitting, we get

$\rm :\longmapsto\:I \: = \: \displaystyle\int\sf \: \dfrac{(x + cosx)}{x(x + cosx)} \: dx \: - \: \displaystyle\int\sf \: \dfrac{x(1 - sinx)}{x(x + cosx)} \: dx$

$\rm :\longmapsto\:I \: = \: \displaystyle\int\sf \: \dfrac{1}{x} \: dx \: - \: \displaystyle\int\sf \: \dfrac{1 - sinx}{x + cosx} \: dx$

We know,

$\red{ \boxed{ \bf \: \displaystyle\int\sf \: \frac{1}{x}dx = log(x) + c}}$

$\red{ \boxed{ \bf \: \displaystyle\int\sf \: \frac{f'(x)}{f(x)}dx = log(f(x)) + c}}$

So, apply these results, we get

$\rm :\implies\:I \: = \: log(x) - log(x + cosx) + c$

$\red{\bigg \{ \because \: \dfrac{d}{dx}(x + cosx) = 1 - sinx \bigg \}}$

$\bf :\implies\:I = log \: \bigg(\dfrac{x}{x + cosx} \bigg) + c$

Additional Information :-

$\rm :\longmapsto\:\displaystyle\int\sf \: cosx \: dx \: = \: sinx \: + \: c$

$\rm :\longmapsto\:\displaystyle\int\sf \: sinx \: dx \: = - \: cosx \: + \: c$

$\rm :\longmapsto\:\displaystyle\int\sf \: secx \: dx \: = \: log(secx + tanx)\: + \: c$

$\rm :\longmapsto\:\displaystyle\int\sf \: cosecx \: dx \: = \: log(cosecx - cotx)\: + \: c$

$\rm :\longmapsto\:\displaystyle\int\sf \: tanx \: dx \: = \: log(secx)\: + \: c$

$\rm :\longmapsto\:\displaystyle\int\sf \: cotx \: dx \: = \: log(sinx)\: + \: c$