Integral Calculus
86283
If \(\int e^{x}\left(\frac{x+2}{x+4}\right)^{2} d x=f(x)+\) arbitrary constant, then \(\mathbf{f}(\mathbf{x})=\)
1 \(\frac{x e^{x}}{x+4}\)
2 \(\frac{e^{x}}{x+4}\)
3 \(\frac{\mathrm{xe}^{\mathrm{x}}}{(\mathrm{x}+4)^{2}}\)
4 \(\frac{e^{x}}{(x+4)^{2}}\)
Explanation:
(A) : Given,
\(\int \mathrm{e}^{\mathrm{x}}\left(\frac{\mathrm{x}+2}{\mathrm{x}+4}\right)^{2} \mathrm{dx}=\mathrm{f}(\mathrm{x})+\) arbitrary constant
\(=\int \mathrm{e}^{x}\left(\frac{\mathrm{x}^{2}+4+4 \mathrm{x}}{(\mathrm{x}+4)^{2}}\right) \mathrm{dx}\)
\(=\int \mathrm{e}^{\mathrm{x}}\left(\frac{\mathrm{x}}{\mathrm{x}+4}+\frac{4}{(\mathrm{x}+4)^{2}}\right) \mathrm{dx}\)
Let, \(\quad g(x)=\frac{x}{x+4}\)
\(g^{\prime}(x)=\frac{4}{(x+4)^{2}}\)
So,
\(=\int \mathrm{e}^{\mathrm{x}}\left\{\mathrm{g}(\mathrm{x})+\mathrm{g}^{\prime}(\mathrm{x})\right\} \mathrm{dx}=\mathrm{e}^{\mathrm{x}} \mathrm{g}(\mathrm{x})+\mathrm{C}\)
\(=\mathrm{e}^{\mathrm{x}}\left(\frac{\mathrm{x}}{\mathrm{x}+4}\right)+\mathrm{C}\)
\(\therefore \quad \mathrm{f}(\mathrm{x})=\frac{\mathrm{x} \cdot \mathrm{e}^{\mathrm{x}}}{\mathrm{x}+4}+\mathrm{C}\)