If $$f(x)=$$∫$$5x4+4x5x5+x+12dx$$ and $$f(0)=0$$, then $$f(1)=$$

$$fx=$$∫$$5x4+4x5x5+x+12dx=$$∫$$5x2+4x5x101+1x4+1x52=$$∫$$5x6+4x51+1x4+1x52dxPut1+1x4+1x5=t$$⇒$$-4x5-5x6dx=ct$$∫−$$1t2dt=1t+Cfx=11+1x4+1x5+C=x5x5+x+1+Cfx=x5x5+x+1f0=0+C$$⇒$$0+C$$⇒$$C=0f1=13=0.33$$

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$If f (x) = \int \frac{5 x^{4} + 4 x^{5}}{{(x^{5} + x + 1)}^{2}} d x and f (0) = 0, then f (1) =$

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Solution

$f (x) = \int \frac{5 x^{4} + 4 x^{5}}{{(x^{5} + x + 1)}^{2}} d x = \int \frac{5 x^{2} + 4 x^{5}}{x^{10} {(1 + \frac{1}{x^{4}} + \frac{1}{x^{5}})}^{2}}$
$= \int \frac{\frac{5}{x^{6}} + \frac{4}{x^{5}}}{{(1 + \frac{1}{x^{4}} + \frac{1}{x^{5}})}^{2}} d x [\begin{array}{l} P u t 1 + \frac{1}{x^{4}} + \frac{1}{x^{5}} = t \\ \Rightarrow (\frac{- 4}{x^{5}} - \frac{5}{x^{6}}) d x = c t \end{array}]$
$\int - \frac{1}{t^{2}} d t = \frac{1}{t} + C$
$f (x) = \frac{1}{(1 + \frac{1}{x^{4}} + \frac{1}{x^{5}})} + C = \frac{x^{5}}{(x^{5} + x + 1)} + C$
$f (x) = \frac{x^{5}}{x^{5} + x + 1} [f (0) = 0 + C \Rightarrow 0 + C \Rightarrow C = 0]$
$f (1) = \frac{1}{3} = 0.33$

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Methods of integration

Remember concepts with our Masterclasses.

If f(x)=∫5x4+4x5x5+x+12dx and f(0)=0, then f(1)=

fx=∫5x4+4x5x5+x+12dx=∫5x2+4x5x101+1x4+1x52=∫5x6+4x51+1x4+1x52dxPut1+1x4+1x5=t⇒-4x5-5x6dx=ct∫−1t2dt=1t+Cfx=11+1x4+1x5+C=x5x5+x+1+Cfx=x5x5+x+1f0=0+C⇒0+C⇒C=0f1=13=0.33

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$If f (x) = \int \frac{5 x^{4} + 4 x^{5}}{{(x^{5} + x + 1)}^{2}} d x and f (0) = 0, then f (1) =$