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cx-b-2-2x-cx-2b-2-4x-cx-2-4b-2-8x-dx-




Question Number 185837 by test1234 last updated on 28/Jan/23
∫((cx+b^2 )/(2x))−((cx+2b^2 )/(4x))+(((cx)^2 +4b^2 )/(8x))dx=?
$$\int\frac{{cx}+{b}^{\mathrm{2}} }{\mathrm{2}{x}}−\frac{{cx}+\mathrm{2}{b}^{\mathrm{2}} }{\mathrm{4}{x}}+\frac{\left({cx}\right)^{\mathrm{2}} +\mathrm{4}{b}^{\mathrm{2}} }{\mathrm{8}{x}}{dx}=? \\ $$
Commented by MJS_new last updated on 28/Jan/23
seriously?  =(c^2 /8)∫xdx+(c/4)∫dx+(b^2 /2)∫(dx/x)=  =(c^2 /(16))x^2 +(c/4)x+(b^2 /2)ln ∣x∣ +C
$$\mathrm{seriously}? \\ $$$$=\frac{{c}^{\mathrm{2}} }{\mathrm{8}}\int{xdx}+\frac{{c}}{\mathrm{4}}\int{dx}+\frac{{b}^{\mathrm{2}} }{\mathrm{2}}\int\frac{{dx}}{{x}}= \\ $$$$=\frac{{c}^{\mathrm{2}} }{\mathrm{16}}{x}^{\mathrm{2}} +\frac{{c}}{\mathrm{4}}{x}+\frac{{b}^{\mathrm{2}} }{\mathrm{2}}\mathrm{ln}\:\mid{x}\mid\:+{C} \\ $$

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