Solution assignment 06 Integration by substitution

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Assignment 6

Solve:

$\displaystyle\int{\displaystyle\frac{2x+1}{4x^2-1}}dx$

Solution

This integral looks difficult, but when we have factorized the denominator, we will find that the integral is in fact much simpler (of course we have $x\neq{\displaystyle\frac{1}{2}}$ and $x\neq{-\displaystyle\frac{1}{2}}$ ):

$\displaystyle\frac{2x+1}{4x^2-1}=\displaystyle\frac{2x+1}{(2x+1)(2x-1)}=\displaystyle\frac{1}{2x-1}$

The integral becomes:

$\displaystyle\int{\displaystyle\frac{2x+1}{4x^2-1}}dx=\displaystyle\int{\displaystyle\frac{1}{2x-1}}dx$

The primitive is:

$\displaystyle\frac{1}{2}\ln(|2x-1|)$

and thus:

$\displaystyle\int{\displaystyle\frac{1}{2x-1}}dx=\displaystyle\frac{1}{2}\ln(|2x-1|)+C$

We can verify the result by differentiation. The chain rule has to be applied.

Return to Assignments Integration by substitution

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Solution assignment 06 Integration by substitution

Assignment 6

Solution

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