If I = ( x 2 + 1 ) ( ( x + 1 ) e x ) 2 d x = A ( f ( x ) ) 2 + C , where C is constant of integration and f ( - 1 ) = 2 e , then 2 A + f ( 0 ) is

Question

If $I = \int (x^{2} + 1) ({(x + 1) e^{x})}^{2} d x = A (f {(x))}^{2} + C,$ where C is constant of integration and $f (- 1) = \frac{2}{e},$ then $2 A + f (0)$ is

Smart Achievers · Accepted Answer

(2)

$I = \int (x^{2} + 1) ({(x + 1) e^{x})}^{2} d x$

$(x^{2} + 1) e^{x} = t$

$\Rightarrow {(x + 1)}^{2} e^{x} d x = d t$

$I = \frac{1}{2} ({(x^{2} + 1) e^{x})}^{2} + c$

$= \frac{1}{2} (f {(x))}^{2} + c$

$2 A + f (0) = 1 + 1 = 2$

Solution

Q.
If $I = \int (x^{2} + 1) ({(x + 1) e^{x})}^{2} d x = A (f {(x))}^{2} + C,$ where C is constant of integration and $f (- 1) = \frac{2}{e},$ then $2 A + f (0)$ is

Ans.
(2)

$I = \int (x^{2} + 1) ({(x + 1) e^{x})}^{2} d x$

$(x^{2} + 1) e^{x} = t$

$\Rightarrow {(x + 1)}^{2} e^{x} d x = d t$

$I = \frac{1}{2} ({(x^{2} + 1) e^{x})}^{2} + c$

$= \frac{1}{2} (f {(x))}^{2} + c$

$2 A + f (0) = 1 + 1 = 2$

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Solution

Q. If I=∫(x2+1)((x+1)ex)2dx=A(f(x))2+C, where C is constant of integration and f(-1)=2e, then 2A+f(0) is

Ans. (2) I=∫(x2+1)((x+1)ex)2dx (x2+1)ex=t ⇒(x+1)2exdx=dt I=12((x2+1)ex)2+c =12(f(x))2+c 2A+f(0)=1+1=2

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Q.
If $I = \int (x^{2} + 1) ({(x + 1) e^{x})}^{2} d x = A (f {(x))}^{2} + C,$ where C is constant of integration and $f (- 1) = \frac{2}{e},$ then $2 A + f (0)$ is

Ans.
(2)

$I = \int (x^{2} + 1) ({(x + 1) e^{x})}^{2} d x$

$(x^{2} + 1) e^{x} = t$

$\Rightarrow {(x + 1)}^{2} e^{x} d x = d t$

$I = \frac{1}{2} ({(x^{2} + 1) e^{x})}^{2} + c$

$= \frac{1}{2} (f {(x))}^{2} + c$

$2 A + f (0) = 1 + 1 = 2$