If ∫x3ex2dx=eK.l+c then K,l=

I=∫x3ex2dxx2=t⇒2x dx=dtI=12∫t.et.dt=et2(t−1)=ex22(x2−1)+c∴(k,l)=x2,x2−12

If ∫x3ex2dx=eK.l+c then K,l=

I=∫x3ex2dxx2=t⇒2x dx=dtI=12∫t.et.dt=et2(t−1)=ex22(x2−1)+c∴(k,l)=x2,x2−12

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If $\int x^{3} e^{x^{2}} d x = e^{K} . l + c$ then $K, l =$

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$x^{2}, 2 x^{2} - 1$

$x^{2}, \frac{x^{2} - 1}{2}$

$x, \frac{x^{2} - 1}{2}$

$x^{2}, x^{2} - 2$

answer is D.

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Detailed Solution

$I = \int x^{3} e^{x^{2}} d x$

$x^{2} = t$

$\Rightarrow 2 x d x = d t$

$\begin{array}{l} I = \frac{1}{2} \int t . e^{t} . d t \\ = \frac{e^{t}}{2} (t - 1) \\ = \frac{e^{x^{2}}}{2} (x^{2} - 1) + c \\ ∴ (k, l) = (x^{2}, \frac{x^{2} - 1}{2}) \end{array}$

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If $\int x^{3} e^{x^{2}} d x = e^{K} . l + c$ then $K, l =$