Question-201452

Question Number 201452 by tri26112004 last updated on 06/Dec/23

Answered by Calculusboy last updated on 06/Dec/23

∫x^(−2) e^(−4x) dx Solution: by using IBP let u=e^(−4x) du=−4e^(−4x) dx dv=x^(−2) v=−x^(−1) I=uv−∫vdudx I=−x^(−1) e^(−4x) −4∫x^(−1) e^(−4x) dx (take Ibp_2 ) I_1 =∫x^(−1) e^(−4x) dx let u=e^(−4x) du=−4e^(−4x) dx dv=x^(−1) v=(x^(−1+1) /(−1+1)) I_1 =uv−∫vdudx I_1 =0 then I=−x^(−1) e^(−4x) −4I_1 I=−x^(−1) e^(−4x) +C

$$\int\boldsymbol{{x}}^{−\mathrm{2}} \boldsymbol{{e}}^{−\mathrm{4}\boldsymbol{{x}}} \boldsymbol{{dx}} \\ $$$$\boldsymbol{{Solution}}:\:\:\boldsymbol{{by}}\:\boldsymbol{{using}}\:\boldsymbol{{IBP}} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{u}}=\boldsymbol{{e}}^{−\mathrm{4}\boldsymbol{{x}}} \:\:\:\boldsymbol{{du}}=−\mathrm{4}\boldsymbol{{e}}^{−\mathrm{4}\boldsymbol{{x}}} \boldsymbol{{dx}}\:\:\boldsymbol{{dv}}=\boldsymbol{{x}}^{−\mathrm{2}} \:\:\:\boldsymbol{{v}}=−\boldsymbol{{x}}^{−\mathrm{1}} \\ $$$$\boldsymbol{{I}}=\boldsymbol{{uv}}−\int\boldsymbol{{vdudx}} \\ $$$$\boldsymbol{{I}}=−\boldsymbol{{x}}^{−\mathrm{1}} \boldsymbol{{e}}^{−\mathrm{4}\boldsymbol{{x}}} −\mathrm{4}\int\boldsymbol{{x}}^{−\mathrm{1}} \boldsymbol{{e}}^{−\mathrm{4}\boldsymbol{{x}}} \boldsymbol{{dx}}\:\:\:\left(\boldsymbol{{take}}\:\boldsymbol{{Ibp}}_{\mathrm{2}} \right) \\ $$$$\boldsymbol{{I}}_{\mathrm{1}} =\int\boldsymbol{{x}}^{−\mathrm{1}} \boldsymbol{{e}}^{−\mathrm{4}\boldsymbol{{x}}} \boldsymbol{{dx}}\:\:\:\boldsymbol{{let}}\:\boldsymbol{{u}}=\boldsymbol{{e}}^{−\mathrm{4}\boldsymbol{{x}}} \:\:\boldsymbol{{du}}=−\mathrm{4}\boldsymbol{{e}}^{−\mathrm{4}\boldsymbol{{x}}} \boldsymbol{{dx}}\:\:\boldsymbol{{dv}}=\boldsymbol{{x}}^{−\mathrm{1}} \:\:\:\boldsymbol{{v}}=\frac{{x}^{−\mathrm{1}+\mathrm{1}} }{−\mathrm{1}+\mathrm{1}} \\ $$$$\boldsymbol{{I}}_{\mathrm{1}} =\boldsymbol{{uv}}−\int\boldsymbol{{vdudx}} \\ $$$$\boldsymbol{{I}}_{\mathrm{1}} =\mathrm{0}\:\:\:\boldsymbol{{then}} \\ $$$$\boldsymbol{{I}}=−\boldsymbol{{x}}^{−\mathrm{1}} \boldsymbol{{e}}^{−\mathrm{4}\boldsymbol{{x}}} −\mathrm{4}\boldsymbol{{I}}_{\mathrm{1}} \\ $$$$\boldsymbol{{I}}=−\boldsymbol{{x}}^{−\mathrm{1}} \boldsymbol{{e}}^{−\mathrm{4}\boldsymbol{{x}}} +\boldsymbol{{C}} \\ $$

Commented by Calculusboy last updated on 06/Dec/23

$$\boldsymbol{{but}}\:\boldsymbol{{am}}\:\boldsymbol{{not}}\:\boldsymbol{{sure}}\:\boldsymbol{{sha}} \\ $$

Commented by Frix last updated on 07/Dec/23

$${dv}={x}^{−\mathrm{1}} \:\Rightarrow\:{v}=\mathrm{ln}\:{x} \\ $$

Commented by Calculusboy last updated on 08/Dec/23

$$\boldsymbol{{oh}}\:\boldsymbol{{thanks}}\:\boldsymbol{{for}}\:\boldsymbol{{reminding}}\:\boldsymbol{{me}}\:\boldsymbol{{sir}} \\ $$

Answered by Mathspace last updated on 07/Dec/23

I=∫ (e^(−4x) /x^2 )dx by parts I=−(1/x)e^(−4x) −∫(−(1/x))(−4)e^(−4x) dx =−(e^(−4x) /x)−∫ (e^(−4x) /x)dx =−(e^(−4x) /x)−∫(1/x)Σ_(n=0) ^∞ (((−4x)^n )/(n!))dx =−(e^(−4x) /x)−Σ_(n=0) ^∞ (((−4)^n )/(n!))∫x^(n−1) dx =−(e^(−4x) /x)−Σ_(n=0) ^∞ (((−4)^n )/(n(n!))) x^n

$${I}=\int\:\frac{{e}^{−\mathrm{4}{x}} }{{x}^{\mathrm{2}} }{dx}\:{by}\:{parts} \\ $$$${I}=−\frac{\mathrm{1}}{{x}}{e}^{−\mathrm{4}{x}} −\int\left(−\frac{\mathrm{1}}{{x}}\right)\left(−\mathrm{4}\right){e}^{−\mathrm{4}{x}} {dx} \\ $$$$=−\frac{{e}^{−\mathrm{4}{x}} }{{x}}−\int\:\frac{{e}^{−\mathrm{4}{x}} }{{x}}{dx} \\ $$$$=−\frac{{e}^{−\mathrm{4}{x}} }{{x}}−\int\frac{\mathrm{1}}{{x}}\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{4}{x}\right)^{{n}} }{{n}!}{dx} \\ $$$$=−\frac{{e}^{−\mathrm{4}{x}} }{{x}}−\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{4}\right)^{{n}} }{{n}!}\int{x}^{{n}−\mathrm{1}} {dx} \\ $$$$=−\frac{{e}^{−\mathrm{4}{x}} }{{x}}−\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{4}\right)^{{n}} }{{n}\left({n}!\right)}\:{x}^{{n}} \\ $$

Commented by Calculusboy last updated on 11/Dec/23

sir,i think you forget the 4 you can check your solution back

$$\boldsymbol{{sir}},\boldsymbol{{i}}\:\boldsymbol{{think}}\:\boldsymbol{{you}}\:\boldsymbol{{forget}}\:\boldsymbol{{the}}\:\mathrm{4} \\ $$$$\boldsymbol{{you}}\:\boldsymbol{{can}}\:\boldsymbol{{check}}\:\boldsymbol{{your}}\:\boldsymbol{{solution}}\:\boldsymbol{{back}} \\ $$

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