*u* = 1+3*e*^{-x}

so that (Don't forget to use the chain rule on *e*^{-x}.)

*du* = 3*e*^{-x}(-1) *dx* = -3*e*^{-x} *dx* ,

or

(-1/3)*du* = *e*^{-x} *dx* .

However, how can we replace the term *e*^{-3x} in the original problem ? Note that

.

From the u-substitution

*u* = 1+3*e*^{-x} ,

we can "back substitute" with

*e*^{-x} = (1/3)(*u*-1) .

Substitute into the original problem, replacing all forms of *x*, getting

(Recall that
(*AB*)^{C} = *A*^{C} *B*^{C} .)

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* SOLUTION 11 :* Integrate
. Use u-substitution. Let

*u* = *e*^{2x}+6*e*^{x}+ 1

so that (Don't forget to use the chain rule on *e*^{2x}.)

*du* = (2*e*^{2x}+6*e*^{x}) *dx*

= (2*e*^{x+x}+6*e*^{x}) *dx*

= (2*e*^{x}*e*^{x}+6*e*^{x}) *dx*

= 2*e*^{x}(*e*^{x}+3) *dx*

= 2*e*^{x}(3+*e*^{x}) *dx*

or

(1/2) *du* = *e*^{x}(3+*e*^{x}) *dx* .

Substitute into the original problem, replacing all forms of *x*, getting

(Do not make the following VERY COMMON MISTAKE : . Why is this INCORRECT ?)

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* SOLUTION 12 :* Integrate
.
First, factor out *e*^{9x} from inside the parantheses. Then

(Recall that
(*AB*)^{C} = *A*^{C} *B*^{C} .)

(Recall that
(*A*^{B})^{C} = *A*^{BC} .)

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Now use u-substitution. Let

*u* = 27+*e*^{3x}

so that (Don't forget to use the chain rule on *e*^{3x}.)

*du* = 3*e*^{3x} *dx* ,

or

(1/3) *du* = *e*^{3x} *dx* .

Substitute into the original problem, replacing all forms of *x* , and getting

.

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