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

Answered by Calculusboy last updated on 06/Dec/23

$$\int {\mathit{x}}^{-2}{\mathit{e}}^{-4\mathit{x}}\mathit{d}\mathit{x}$$$$\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}:\mathit{b}\mathit{y}\mathit{u}\mathit{s}\mathit{i}\mathit{n}\mathit{g}\mathit{I}\mathit{B}\mathit{P}$$$$\mathit{l}\mathit{e}\mathit{t}\mathit{u}={\mathit{e}}^{-4\mathit{x}}\mathit{d}\mathit{u}=-4{\mathit{e}}^{-4\mathit{x}}\mathit{d}\mathit{x}\mathit{d}\mathit{v}={\mathit{x}}^{-2}\mathit{v}=-{\mathit{x}}^{-1}$$$$\mathit{I}=\mathit{u}\mathit{v}-\int \mathit{v}\mathit{d}\mathit{u}\mathit{d}\mathit{x}$$$$\mathit{I}=-{\mathit{x}}^{-1}{\mathit{e}}^{-4\mathit{x}}-4\int {\mathit{x}}^{-1}{\mathit{e}}^{-4\mathit{x}}\mathit{d}\mathit{x}\left(\mathit{t}\mathit{a}\mathit{k}\mathit{e}{\mathit{I}\mathit{b}\mathit{p}}_2\right)$$$${\mathit{I}}_1=\int {\mathit{x}}^{-1}{\mathit{e}}^{-4\mathit{x}}\mathit{d}\mathit{x}\mathit{l}\mathit{e}\mathit{t}\mathit{u}={\mathit{e}}^{-4\mathit{x}}\mathit{d}\mathit{u}=-4{\mathit{e}}^{-4\mathit{x}}\mathit{d}\mathit{x}\mathit{d}\mathit{v}={\mathit{x}}^{-1}\mathit{v}=\frac{x^{-1+1}}{-1+1}$$$${\mathit{I}}_1=\mathit{u}\mathit{v}-\int \mathit{v}\mathit{d}\mathit{u}\mathit{d}\mathit{x}$$$${\mathit{I}}_1=0\mathit{t}\mathit{h}\mathit{e}\mathit{n}$$$$\mathit{I}=-{\mathit{x}}^{-1}{\mathit{e}}^{-4\mathit{x}}-4{\mathit{I}}_1$$$$\mathit{I}=-{\mathit{x}}^{-1}{\mathit{e}}^{-4\mathit{x}}+\mathit{C}$$

Commented by Calculusboy last updated on 06/Dec/23

$$\mathit{b}\mathit{u}\mathit{t}\mathit{a}\mathit{m}\mathit{n}\mathit{o}\mathit{t}\mathit{s}\mathit{u}\mathit{r}\mathit{e}\mathit{s}\mathit{h}\mathit{a}$$

Commented by Frix last updated on 07/Dec/23

$$dv=x^{-1}\Rightarrow v=\ln x$$

Commented by Calculusboy last updated on 08/Dec/23

$$\mathit{o}\mathit{h}\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{f}\mathit{o}\mathit{r}\mathit{r}\mathit{e}\mathit{m}\mathit{i}\mathit{n}\mathit{d}\mathit{i}\mathit{n}\mathit{g}\mathit{m}\mathit{e}\mathit{s}\mathit{i}\mathit{r}$$

Answered by Mathspace last updated on 07/Dec/23

$$I=\int \frac{e^{-4x}}{x^2}dxbyparts$$$$I=-\frac{1}{x}{e}^{-4x}-\int (-\frac{1}{x})(-4)e^{-4x}dx$$$$=-\frac{e^{-4x}}{x}-\int \frac{e^{-4x}}{x}dx$$$$=-\frac{e^{-4x}}{x}-\int \frac{1}{x}\sum _{n=0}^{\mathrm{\infty }}\frac{{(-4x)}^n}{n!}dx$$$$=-\frac{e^{-4x}}{x}-\sum _{n=0}^{\mathrm{\infty }}\frac{{(-4)}^n}{n!}\int x^{n-1}dx$$$$=-\frac{e^{-4x}}{x}-\sum _{n=0}^{\mathrm{\infty }}\frac{{(-4)}^n}{n(n!)}{x}^n$$

Commented by Calculusboy last updated on 11/Dec/23

$$\mathit{s}\mathit{i}\mathit{r},\mathit{i}\mathit{t}\mathit{h}\mathit{i}\mathit{n}\mathit{k}\mathit{y}\mathit{o}\mathit{u}\mathit{f}\mathit{o}\mathit{r}\mathit{g}\mathit{e}\mathit{t}\mathit{t}\mathit{h}\mathit{e}4$$$$\mathit{y}\mathit{o}\mathit{u}\mathit{c}\mathit{a}\mathit{n}\mathit{c}\mathit{h}\mathit{e}\mathit{c}\mathit{k}\mathit{y}\mathit{o}\mathit{u}\mathit{r}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{b}\mathit{a}\mathit{c}\mathit{k}$$

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