Calculate.
-$\int _ { 0 } ^ { 1 } \left( e ^ { 3 x } - \frac { 1 } { ( x + 1 ) ^ { 2 } } \right) d x$
A) $\frac { 1 } { 3 }$ $\mathrm { e } ^ { 3 }$ + $\frac { 7 } { 6 }$
B) $\frac { 1 } { 3 }$ e + $\frac { 7 } { 6 }$
C) $\frac { 1 } { 3 }$ $\mathrm { e } ^ { 3 }$ - e - $\frac { 1 } { 2 }$
D) $\frac { 1 } { 3 } \mathrm { e } ^ { 3 } - \frac { 5 } { 6 }$
E) none of these

Question

Quizplus · Accepted Answer

The integral can be split into two parts and solved separately. The first part, $\int_0^1 e^{3x} dx$, integrates to $\frac{1}{3}e^{3x}$ evaluated from 0 to 1, which gives $\frac{1}{3}e^3 - \frac{1}{3}$. The second part, $\int_0^1 -\frac{1}{(x+1)^2} dx$, integrates to $ \frac{1}{x+1}$ evaluated from 0 to 1, which gives 1 - $\frac{1}{2}$ = $\frac{1}{2}$. Combining these results gives $\frac{1}{3}e^3 - \frac{1}{3} - \frac{1}{2}$, simplifying to $\frac{1}{3}e^3 - \frac{5}{6}$.

Calculate - ∫01(e3x−1(x+1)2)dx\int _ { 0 } ^ { 1 } \left( e ^ { 3 x } - \frac { 1 } { ( x + 1 ) ^ { 2 } } \right) d x∫01​(e3x−(x+1)21​)dx

Calculate
- $\int _ { 0 } ^ { 1 } \left( e ^ { 3 x } - \frac { 1 } { ( x + 1 ) ^ { 2 } } \right) d x$