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x-1-x-dx-




Question Number 5265 by Kasih last updated on 03/May/16
∫ (x/(1+x)) dx
$$\int\:\frac{{x}}{\mathrm{1}+{x}}\:{dx} \\ $$
Answered by Yozzii last updated on 03/May/16
(x/(1+x))=1−(1/(1+x))=((1+x−1)/(1+x))=(x/(1+x))  ⇒∫(x/(1+x))dx=∫(1−(1/(1+x)))dx=x−ln∣x+1∣+C
$$\frac{{x}}{\mathrm{1}+{x}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{x}}=\frac{\mathrm{1}+{x}−\mathrm{1}}{\mathrm{1}+{x}}=\frac{{x}}{\mathrm{1}+{x}} \\ $$$$\Rightarrow\int\frac{{x}}{\mathrm{1}+{x}}{dx}=\int\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{x}}\right){dx}={x}−{ln}\mid{x}+\mathrm{1}\mid+{C} \\ $$$$ \\ $$
Commented by Kasih last updated on 09/May/16
can it use integration by parts?
$${can}\:{it}\:{use}\:{integration}\:{by}\:{parts}? \\ $$
Commented by 1771727373 last updated on 14/May/16
no you cant
$${no}\:{you}\:{cant} \\ $$

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