Question

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

Answer 1

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`

Answer 2

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`.