Consider the following integral.

Consider the following integral.
Find a substitution to rewrite the integrand as

.

Evaluate the given integral. (Use C for the constant of integration. Remember to use absolute values where appropriate.)

The Correct Answer and Explanation is:

Answer:
The required substitution is 

u=ex+6u=ex+6

, which rewrites the integrand as 

1u2+1u2+11​

.

The value of the integral is:

∫ex(e2x)+12ex+37dx=arctan⁡(ex+6)+C∫(e2x)+12ex+37ex​dx=arctan(ex+6)+C

Explanation:

The given integral is 

∫ex(e2x)+12ex+37dx∫(e2x)+12ex+37ex​dx

.

To evaluate this integral, an appropriate substitution is the first step. Observing the numerator, 

exex

, which is the derivative of 

exex

, suggests a substitution involving 

exex

. The denominator is a quadratic in terms of 

exex

, since 

e2x=(ex)2e2x=(ex)2

. This structure points towards a method involving completing the square followed by an arctangent integration formula.

Let’s begin by focusing on the denominator: 

e2x+12ex+37e2x+12ex+37

. To complete the square for the terms involving 

exex

, one takes half of the coefficient of 

exex

 and squares it. The coefficient is 12, half of which is 6, and its square is 36. The denominator can be rewritten by adding and subtracting 36:

(e2x+12ex+36)−36+37(e2x+12ex+36)−36+37

The expression in the parenthesis is a perfect square:

(ex+6)2+1(ex+6)2+1

Now, the integral becomes:

∫ex(ex+6)2+1dx∫(ex+6)2+1ex​dx

This form suggests the substitution 

u=ex+6u=ex+6

. The differential of 

uu

 is found by differentiating with respect to 

xx

:

dudx=ex⟹du=exdxdxdu​=ex⟹du=exdx

This substitution perfectly transforms the integral. The numerator, 

exdxexdx

, becomes 

dudu

, and the denominator, 

(ex+6)2+1(ex+6)2+1

, becomes 

u2+1u2+1

. The integrand is now rewritten as 

1u2+1u2+11​

.

The integral in terms of 

uu

 is:

∫1u2+1du∫u2+11​du

This is a standard integral form whose antiderivative is the arctangent function. Specifically, 

∫1u2+a2du=1aarctan⁡(ua)+C∫u2+a21​du=a1​arctan(au​)+C

. For this problem, 

a=1a=1

.

∫1u2+1du=arctan⁡(u)+C∫u2+11​du=arctan(u)+C

The final step is to substitute back in terms of the original variable, 

xx

. Replacing 

uu

 with 

ex+6ex+6

 gives the final result.

arctan⁡(ex+6)+Carctan(ex+6)+C