Advanced Integration Techniques. a. Evaluate \int^e_1 \frac{\ln x}{x} dx. b. Evaluate the...

Question:

Advanced Integration Techniques.

a. Evaluate {eq}\int^e_1 \frac{\ln x}{x} dx. {/eq}

b. Evaluate the integral using integration by parts.

{eq}\int (2x + 1)e^{-3x} dx {/eq}

c. Evaluate the integral using partial fraction decomposition.

{eq}\int \frac{x^2 - 2x - 2}{(x-1)(x^2 + 2)} dx {/eq}

Advanced Integration Techniques.

In advanced integration techniques u-substitution, integration by parts, partial fraction, Laplace transformation all these methods come under advance integration.

It defines both definite and indefinite integrals.

The formula is:

{eq}\displaystyle\int x^n\ dx=\dfrac{x^{n+1}}{n+1}+c\\\\ \displaystyle\int \dfrac{1}{x}dx=ln\left | x \right |+c\\\\ \displaystyle\int e^x\ dx=e^x+c\\\\ {/eq}

Answer and Explanation:

Here we have to solve the given integration by advanced method of integration:

Part A.)

{eq}\displaystyle\int _1^e\dfrac{lnx}{x}dx\\\\ {/eq}

Substituting for making it possible:

{eq}lnx=z\\\\ \dfrac{1}{x}dx=dz\\\\ {/eq}

Substituting for the values of limits:

Lower limit:

When {eq}x=e\\\\ {/eq}

{eq}z=lne\\\\ z=1\\\\ {/eq}

Upper limit:

When {eq}x=1\\\\ {/eq}

{eq}z=ln(1)\\\\ z=0\\\\ {/eq}

Such as the domain of the integration becomes:

{eq}0\leq z\leq 1\\\\ {/eq}

Hence integrating:

{eq}=\displaystyle\int _0^1z\ dz\\\\ =\left [ \dfrac{z^2}{2} \right ]_0^1\\\\ =\dfrac{1}{2}\left [ 1-0 \right ]\\\\ =\dfrac{1}{2}\\\\ {/eq}

Part B.)

{eq}\displaystyle\int (2x+1)e^{-3x}dx\\\\ {/eq}

Given integration will be evaluated by applying the formula of integration by parts:

{eq}\displaystyle\int (uv)dx=u\displaystyle\int v\ dx-\displaystyle\int \left [ \dfrac{d}{dx}(u)\displaystyle\int v\ dx \right ]dx\\\\ =(2x+1)\displaystyle\int e^{-3x}dx-\displaystyle\int \left [ \dfrac{d}{dx}(2x+1)\displaystyle\int e^{-3x}\ dx \right ]dx\\\\ =(2x+1)\left [ e^{-3x}\left ( \dfrac{-1}{3} \right ) \right ]-\displaystyle\int \left [ 2\cdot e^{-3x}\cdot \left ( \dfrac{-1}{3} \right ) \right ]dx\\\\ =\dfrac{-1}{3}\cdot {e^{-3x}(2x+1)}+\dfrac{2}{3}\displaystyle\int e^{-3x}dx\\\\ =\dfrac{-e^{-3x}(2x+1)}{3}+\dfrac{2}{3}\cdot e^{-3x}\cdot \left ( \dfrac{-1}{3} \right )+c\\\\ =\dfrac{-e^{-3x}(2x+1)}{3}-\dfrac{2}{9}e^{-3x}+c\\\\ {/eq}

Here c defines arbitrary constant.

Part C.)

{eq}\displaystyle\int \dfrac{x^2-2x-2}{(x-1)(x^2+2)}dx\\\\ {/eq}

We can solve the given integration as applying partial fraction decomposition method:

{eq}\dfrac{x^2-2x-2}{(x-1)(x^2+2)}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+2}\\\\ \dfrac{x^2-2x-2}{(x-1)(x^2+2)}=\dfrac{A(x^2+2)+(Bx+C)(x-1)}{(x-1)(x^2+2)}\\\\ x^2-2x-2=x^2A+2A+x^2B-xB+xC-C\\\\ {/eq}

Comparing the coefficients of {eq}x^2 {/eq} form both side of the equation we get:

{eq}A+B=1\,\,\,\,\,eqn(1)\\\\ {/eq}

Comparing the coefficients of {eq}x {/eq} form both side of the equation we get:

{eq}-B+C=-2\,\,\,\,\,eqn(2)\\\\ {/eq}

And by comparing the constant terms we get:

{eq}2A-C=-2\,\,\,\,\,eqn(3)\\\\ {/eq}

Hence solving eqn 1,2 and 3 we get the values:

{eq}A=-1\\\\ B=2\\\\ C=0\\\\ {/eq}

Hence integrating by substituting the values of A,B and C we get:

{eq}=\displaystyle\int \dfrac{-1}{x-1}dx+\displaystyle\int \dfrac{-2x+0}{x^2+2}\\\\ =-ln\left | x-1 \right |-\displaystyle\int \dfrac{2x}{x^2+2}dx\\\\ {/eq}

Let:

{eq}x^2+2=z\\\\ 2x\ dx=dz\\\\ {/eq}

Therefore:

{eq}=-ln\left | x-1 \right |-\displaystyle\int \dfrac{dz}{z}\\\\ =-ln\left | x-1 \right |-ln\left | z \right |+c\\\\ =-ln\left | x-1 \right |-ln\left | x^2+2 \right |+c\\\\ {/eq}

Here c is an arbitrary constant.


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