Integrate \int \frac{x}{2x^2 + 8x + 6} dx 1) using trigonometric substitution; 2) using partial...

Question:

Integrate {eq}\int \frac{x}{2x^2 + 8x + 6} dx {/eq}

1) using trigonometric substitution;

2) using partial fraction decomposition.

Indefinite Integral:

The above problem concerns the topic of the indefinite integral. there are various techniques to solve a indefinite integrals such as

• Direct integration
• Integration by Substitution
• Integration by Partial Fraction
• Integration by Parts

In integration by substitution method, we can assume algebraic function or trigonometric function to integrate. The choice of the function is such that, the derivative of the assumed function is also present in the given integral. After substitution, given integral becomes simple and can be evaluated directly.

In integration by partial fraction method, we decompose the given function into its factor. After that, integration can be evaluated directly.

Formula used for integration:

{eq}\displaystyle \int \csc x dx = \ln |csc x-\cot x|+C = \ln \left | \tan \frac{x}{2} \right | +C {/eq}

Answer and Explanation: 1

Integrate {eq}\displaystyle \int \frac{x}{2x^2 + 8x + 6} dx {/eq}

1) using trigonometric substitution;

{eq}\displaystyle \begin{align} I &= \int \frac{x}{2x^2 + 8x + 6} dx \\ I &= \int \frac{x}{2(x^2+4x+3)}dx \\ I &= \frac{1}{2} \int \frac{x}{x^2+4x+4-4+3} dx\\ I &= \frac{1}{2} \int \frac{x}{x^2+4x+4-4+3} dx\\ I &= \frac{1}{2} \int \frac{x}{(x+2)^2-1 } dx \\ \text{Let } x+2 =\sec \theta &\rightarrow dx = \sec \theta \tan \theta d \theta \\ I &= \frac{1}{2} \int \frac{\sec \theta-2}{\sec ^2 \theta -1}(\sec \theta \tan \theta d \theta ) \\ I &= \frac{1}{2} \int \frac{\sec \theta-2 }{\tan^2 \theta } (\sec \theta \tan \theta d \theta ) \\ I &= \frac{1}{2} \int \left ( \frac{\sec^2 \theta }{\tan \theta } -\frac{2\sec \theta }{\tan \theta } \right )d \theta \\ I &= \frac{1}{2} \int \left ( \frac{1 }{\sin \theta \cos \theta } -\frac{2 }{\sin \theta } \right )d \theta \\ I &= \int \left ( \frac{1 }{2\sin \theta \cos \theta } -\frac{1 }{\sin \theta } \right )d \theta \\ I &= \int \left (\csc 2\theta -\csc \theta \right )d \theta \\ I &= \frac{1}{2} \ln |tan \theta | -\ln | \tan \frac{ \theta}{2} | +C \\ \end{align} {/eq}

Now if {eq}\sec \theta =x+2 \rightarrow \theta = \sec^{-1} (x+2) \text{ and }\tan \theta = \sqrt{x^2+4x+3} {/eq}. Replacing value of {eq}\theta {/eq} and {eq}\tan \theta {/eq} we get

{eq}{\color{Blue} {\displaystyle I= \frac{1}{2} \ln |\sqrt{x^2+4x+3} | -\ln \left | \tan \left ( \frac{ \sec^{-1} (x+2)}{2}\right ) \right | +C }} {/eq}

2) using partial fraction decomposition.

Given integral

{eq}\displaystyle \begin{align} I &= \int \frac{x}{2x^2 + 8x + 6} dx \\ I &= \frac{1}{2} \int \frac{x}{x^2+4x+3}dx \\ I &= \frac{1}{2} \int \frac{x}{(x+1)(x+3)}dx \\ \end{align} {/eq}

Doing partial fraction, we obtain

{eq}\displaystyle \begin{align} \frac{x}{(x+1)(x+3)} &= \frac{A}{x+3}+\frac{B}{x+1}\\ \frac{x}{(x+1)(x+3)} &= \frac{Ax+A+Bx+3B}{(x+1)(x+3)} \\ \end{align} {/eq}

Comparing coefficient, we get

{eq}A+B=1 \\ A+3B= 0 \\ {/eq}

Solving both equation , we get {eq}\displaystyle A=\frac{3}{2} B= \frac{-1}{2}, {/eq}.

Substituting the values of A and b and then integrating, we obtain

{eq}\displaystyle \begin{align} I &= \frac{1}{2} \int \left ( \frac{3}{2(x+3) }- \frac{1}{2(x+1)} \right ) dx \\ I &= \frac{1}{2} \left ( \frac{3}{2} \ln |x+3| - \frac{1}{2} \ln |x+1 | \right )+C \\ I &= {\color{Blue} {\frac{3}{4} \ln |x+3| - \frac{1}{4} \ln |x+1 | +C }} \end{align} {/eq}