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Question Number 136445 by aupo14 last updated on 22/Mar/21

Answered by liberty last updated on 22/Mar/21

$$letf\left(x\right)={\int }_0^x\frac{t-3}{t^2+7}dt$$$$f{}^{\prime }\left(x\right)=\frac{x-3}{x^2+7}=0\Rightarrow x=3$$$$f{}^{″}\left(x\right){\mid }_{x=3}>0\Rightarrow formin$$$$minvalueisf\left(x\right)=\frac{1}{2}\ln \left(10\right)-\arctan 3$$

Commented by aupo14 last updated on 22/Mar/21

can you give me the way you integrate it? im having a bit trouble

Commented by mr W last updated on 22/Mar/21

$$integraliswrong!$$

Commented by liberty last updated on 22/Mar/21

$$hahaha..sorry.i^{\prime }mmisreadthe$$$$integral.ithink$$$${\int }_0^x\frac{t-3}{t^2+1}dt$$

Answered by mr W last updated on 22/Mar/21

$$f\left(x\right)={\int }_0^x\frac{t-3}{t^2+7}dt$$$$={\int }_0^x[\frac{t}{t^2+7}-\frac{3}{t^2+7}]dt$$$$=\frac{1}{2}{\int }_0^x\frac{d(t^2+7)}{t^2+7}-3{\int }_0^x\frac{dt}{t^2+7}$$$$=\frac{1}{2}{\left[\ln (t^2+7)\right]}_0^x-\frac{3}{\sqrt{7}}{\left[{\tan }^{-1}\frac{t}{\sqrt{7}}\right]}_0^x$$$$=\frac{1}{2}\ln \frac{x^2+7}{7}-\frac{3}{\sqrt{7}}{\tan }^{-1}\frac{x}{\sqrt{7}}$$$$f^{\prime }\left(x\right)=\frac{x-3}{x^2+7}=0\Rightarrow x=3$$$$f{\left(x\right)}_{min}=f\left(3\right)=\frac{1}{2}\ln \frac{16}{7}-\frac{3}{\sqrt{7}}{\tan }^{-1}\frac{3}{\sqrt{7}}\approx -0.548$$

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