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.