How can I evaluate this integral? (e^3x - 6/2x+1)dx

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2 Answers
Jan 29, 2018

The answer is:

# int \ (e^{3x}-\frac{6}{2x+1})dx " "=" "\frac{e^{3x}}{3}-3\ln |2x+1|+C#

See the steps below...

Explanation:

# int \ (e^{3x}-\frac{6}{2x+1})dx#

Apply the sum of integral rule,

#int f(x)\pm g(x)dx=\int f(x)dx\pm \int g(x)dx#

#=\int \e^{3x}dx-\int \frac{6}{2x+1}dx#

Solve each integral separately as:

Substitute #u=3x# #""# #\rightarrow# #""# #dx=1/3du#

#=\int \e^u\frac{1}{3}du#

Take the constant out:

#=\frac{1}{3}\cdot \int \e^udu#

Use the integral rule #" "# #\int \e^udu=e^u# #" "# to get:

#=\frac{1}{3}e^u#

Substitute back #u=3x#

#=\frac{1}{3}e^{3x}#

Solve the other integral:

Substitute #u=2x+1# #""# #\rightarrow# #""# #dx=1/2du#

So that after taking the constant out we get:

#=6\cdot \frac{1}{2}\cdot \int \frac{1}{u}du#

Apply the common integral rule #" "# #\int \frac{1}{u}du=\ln (|u|)##" "# to get:

#=3\ln |u|#

By substituting back #u=2x+1#

#=3\ln |2x+1|#

So after adding the constant, the solution becomes:

#=\frac{e^{3x}}{3}-3\ln |2x+1|+C#

Jan 29, 2018

Recall the Rule that,

#intf(x)dx=F(x)+c rArr intf(ax+b)dx=1/aF(ax+b)+c', a ne0.#

It can be very easily proved by using the subst. for #(ax+b)#.

In this accordance, we have,

#inte^xdx=e^x+c_1rArr inte^(3x)dx=1/3e^(3x)+c_1'#, and,

#int1/xdx=ln|x|+c_2rArrint1/(2x+1)dx=1/2ln|(2x+1)|+c_2'#.

Finally, #int(e^(3x)-6/(2x+1))dx#,

#=inte^(3x)dx-int6/(2x+1)dx#,

#=inte^(3x)dx-6int1/(2x+1)dx#,

#=1/3e^(3x)-6*1/2ln|(2x+1)|#,

#=1/3e^(3x)-3ln|(2x+1)|+C#.