The correct option is D e^x(3x^2 5x+6)+c ∫ e^x(3x^2+x+1)dx Lets write 3x^2+x+1=(ax^2+bx+c)+(2ax+b) Comparing the co efficients. a=3; 6+b=1; b ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Geometric Interpretation of Def.Int as Limit of Sum

∫ ex3x2+x+1+c

Question

$\int e^{x} (3 x^{2} + x + 1) + c$

e^{x} (3 x^{2} + 5 x + 1) + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

e^{x} (3 x^{2} + 7 x + 1) + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

e^{x} (3 x^{2} - 5 x + 6) + c

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

e^{x} (3 x^{2} + x) + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $e^{x} (3 x^{2} - 5 x + 6) + c$
$\int e^{x} (3 x^{2} + x + 1) d x$
Lets write
$3 x^{2} + x + 1 = (a x^{2} + b x + c) + (2 a x + b)$
Comparing the co-efficients.
$a = 3; 6 + b = 1; b = - 5, c + b = 1; c = 6$
$3 x^{2} + x + 1 = (3 x^{2} - 5 x + 6) + (6 x - 5)$
$\int e^{x} ([3 x^{2} - 5 x + 6] + (6 x - 5)] d x$
It is in the form of $\int e^{x} (f (x)) + f^{'} (x)] = e^{x} f (x) + c$
So, $\int e^{x} (3 x^{2} + x + 1) d x = e^{x} (3 x^{2} - 5 x + 6) + c$

Suggest Corrections

Similar questions

\int e^{x} (3 x^{2} + 7 x + 2) d x =

\frac{2 x}{{3 x}^{2} - x + 2} - \frac{7 x}{{3 x}^{2} + 5 x + 2} = 1

\frac{- a - \sqrt{b}}{c}

\frac{- a - \sqrt{b}}{c}

b - a - c

\int (1 + 5 x + 10 x^{2} + 10 x^{3} + 5 x^{4} + x^{5}) d x

= \frac{(1 + x)^{p}}{6} + c

p

lim x \to \infty \frac{a (2 x^{3} - x^{2}) + b (x^{3} + 5 x^{2} - 1) - c (3 x^{3} + x^{2})}{a (5 x^{4} - x) - b x^{4} + c (4 x^{4} + 1) + 2 x^{2} + 5 x} = 1

a

b

c

\frac{x^{2} + 5 x + 1}{(x + 1) (x + 2) (x + 3)} = \frac{A}{x + 1} + \frac{B}{(x + 1) (x + 2)} + \frac{C}{(x + 1) (x + 2) (x + 3)}

Join BYJU'S Learning Program