Question

यदि ${\displaystyle {\int }_0^1\frac{{\text{e}}^{\text{t}}}{1+\text{t}}\text{dt}}$ = a, है, तब ${\displaystyle {\int }_0^1\frac{{\text{e}}^{\text{t}}}{{(1+\text{t})}^2}\text{dt}}$  बराबर है

पर्याय

• ${\displaystyle \text{a}-1+\frac{\text{e}}{2}}$

• ${\displaystyle \text{a}+1-\frac{\text{e}}{2}}$

• ${\displaystyle \text{a}-1-\frac{\text{e}}{2}}$

• ${\displaystyle \text{a}+1+\frac{\text{e}}{2}}$

Answer 1

सही उत्तर ${\displaystyle \underline{\text{a}+1-\frac{\text{e}}{2}}}$  है।

व्याख्या:

क्योंकि I = ${\displaystyle {\int }_0^1\frac{{\text{e}}^{\text{t}}}{1+\text{t}}\text{dt}}$

= ${\displaystyle {\left|\frac{1}{1+\text{t}}{\text{e}}^{\text{t}}\right|}_0^1+{\int }_0^1\frac{{\text{e}}^{\text{t}}}{{(1+\text{t})}^2}\text{dt}}$ = a  ...(दिया है)

अत:, ${\displaystyle {\int }_0^1\frac{{\text{e}}^{\text{t}}}{{(1+\text{t})}^2}=\text{a}-\frac{\text{e}}{2}+1}$