Question

Evaluate \(\rm \int (3x^2 +2) e^{(2x^3 +4x)} dx\)
1. \(\rm \frac {1} {2} e^{2x^3 + 4x} + c\)
2. \(\rm e^{2x^3 + 4x} + c\)
3. \(\rm \frac {1} {2} e^{2x^3 + 4} + c\)
4. \(\rm \frac {1} {2} e^{x^3 + 4x} + c\)

Answer 1

Correct Answer - Option 1 : \(\rm \frac {1} {2} e^{2x^3 + 4x} + c\)

Concept:

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm\int {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C

Substitution method: If the function cannot be integrated directly substitution method is used. To integration by substitution is used in the following steps:

• A new variable is to be chosen, say “t”
• The value of dt is to is to be determined.
• Substitution is done and integral function is then integrated.
• Finally, initial variable t, to be returned.

Calculation:

I = \(\rm \int (3x^2 +2) e^{(2x^3 +4x)} dx\)        ----(i)

Let 2x³ + 4x = t

⇒ (6x² + 4) dx = dt

⇒ \(\rm dx = \frac {dt} {6x^2 + 4}\)

Put the value t and dx in the equation (i)

I = \(\rm \int (3x^2 +2) e^t \frac {dt} {6x^2 + 4}\)

= \(\rm \frac {1} {2}\int e^t dt\)

= \(\rm \frac {1} {2}e^t + c\)        ----(ii)      

Put the value of t in equation (ii)

I = \(\rm \frac {1} {2} e^{2x^3 + 4x} + c\)