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Question Number 71175 by 20190927 last updated on 12/Oct/19

$$\int \frac{1}{{\mathrm{x}}^2+2016\mathrm{x}}\mathrm{dx}$$

Commented by Prithwish sen last updated on 12/Oct/19

$$\int \frac{\mathrm{dx}}{{(\mathrm{x}+1008)}^2-{\left(1008\right)}^2}\mathbf{put}\mathbf{x}+1008=\mathbf{t}$$$$\Rightarrow \mathbf{dx}=\mathbf{dt}$$$$\int \frac{\mathbf{dt}}{{\mathbf{t}}^2-{1008}^2}=\frac{1}{2016}\{\int \frac{\mathbf{dt}}{\mathbf{t}-1008}-\int \frac{\mathbf{dt}}{\mathbf{t}+1008}\}$$

Commented by Prithwish sen last updated on 12/Oct/19

$$\mathbf{Another}\mathbf{method}$$$$\int \frac{\mathbf{dx}}{\mathbf{x}(\mathbf{x}+2016)}=\frac{1}{2016}\{\int \frac{\mathbf{dx}}{\mathbf{x}}-\int \frac{\mathbf{dx}}{\mathbf{x}+2016}\}$$

Commented by 20190927 last updated on 13/Oct/19

$$\mathrm{thank}\mathrm{you}$$

Answered by petrochengula last updated on 13/Oct/19

$$=\int \frac{1}{x^2(1+\frac{2016}{x})}dx$$$$lett=1+\frac{2016}{x}\Rightarrow \frac{dt}{dx}=\frac{-2016}{x^2}\Rightarrow \frac{-dt}{2016}=\frac{1}{x^2}dx$$$$=-\frac{1}{2016}\int \frac{dt}{t}=-\frac{1}{2016}ln\mid t\mid +C=-\frac{1}{2016}ln\mid 1+\frac{2016}{x}\mid +C$$

Commented by 20190927 last updated on 13/Oct/19

$$\mathrm{thank}\mathrm{you}$$

Commented by Prithwish sen last updated on 13/Oct/19

$$\mathrm{Amazing}!\mathrm{Sir}.$$

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