Question

∫ (log x+1/x²) eˣ dx=......+c ,Select correct option from the given options.
(a) eˣ (log x+1/x²)
(b) eˣ (log x+1/x)
(c) eˣ (log x-1/x²)
(d) eˣ (log x-1/x)

Answer 1

answer : option (d)

explanation : ∫(logx + 1/x²) e^x dx

= ∫(logx + 1/x - 1/x + 1/x²) e^x dx

= ∫(logx + 1/x)e^x dx - ∫(1/x - 1/x²)e^x dx

from application of integration ,

$\int{e^x(f(x)+f'(x))}\,dx=e^xf(x)+C$

in above question,

if we assume , logx = f(x)

differentiating both sides,

1/x = f'(x)

hence, (logx + 1/x) = {f(x) + f'(x)}

so, $\int{e^x(logx+1/x)}\,dx=e^xlogx$

similarly, let 1/x = g(x)

differentiating both sides,

-1/x² = g'(x)

hence, (1/x - 1/x²) = {g(x) + g'(x)}

so, $\int{e^x(1/x-1/x^2)}\,dx=e^x(1/x)$

hence, ∫(logx + 1/x)e^x dx - ∫(1/x - 1/x²)e^x dx = e^x logx - e^x (1/x) + C

= e^x(logx - 1/x) + C

so, option (d) is correct choice.

Answer 2

• function , f(x) = (x - 1)e^x
• differentiating f(x) with respect to x,

f'(x) = d[(x - 1)e^x + 1]/dx

= d[(x - 1)e^x]/dx + d(1)/dx

= e^x d(x - 1)/dx + (x - 1) d(e^x)/dx + 0

= e^x × 1 + (x - 1) × e^x

= e^x + (x - 1)e^x

= e^x(x - 1 + 1)

= xe^x

hence, f'(x) = xe^x

for all x > 0

⇒e^x > e^0

⇒ e^x > 1 > 0

so, f'(x) = xe^x > 0

$\textsf{we know, any function f is increasing in (a,b) only when f' > 0 in interval (a, b).}$

$\textsf{here, f'(x) > 0 in interval, x > 0 }$

$\textsf{so, f(x) is increasing for all x > 0}$

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Note:-

$\textsf{kindly view answer from brainly .in for better understanding!}$