## Standard Integrals

The page Standard integrals is going to teach formulas in standard integrals and where we need to apply these formulas in integration.

∫ dx/(a² - x²) = (1/2a) log [(a + x)/(a - x)] + c

∫ dx/( - a² ) = (1/2a) log [(x - a)/(x + a)] + c

∫ dx/( a² + ) = (1/a) tan⁻ ¹ (x/a) + c

dx/(a² - x²) = sin⁻ ¹ (x/a) + c

∫ dx/( - a² ) =  log [ x + ( - a² )] + c

∫ dx/( + a² ) =  log [ x + ( + a² )] + c Example 1:

Integrate 1/(1 - 9x²) with respect to x.

This problem can be done in two ways.

Method 1:

Solution:

=   ∫ [1/(1 - 9x²)] dx

=   ∫ dx/(1 - 3²x²)

=   ∫ dx/[ (1 - (3x²) ]

Formula:

∫ dx/(a² - x²) = (1/2a) log [(a + x)/(a - x)] + c

The previous step exactly matches the above formula. In this we have "1" instead of "a" and "3x" instead of "x".

=  {1/2(1) x log [ (1 + 3x)/(1 - 3x) ]  + C} x (1/3)

=  {(1/2) x (1/3) x log [ (1 + 3x)/(1 - 3x) ]  + C}

=  (1/6) x log [ (1 + 3x)/(1 - 3x) ]  + C

Method 2:

Integrate 1/(1 - 9x²) with respect to x.

Solution:

=   ∫ [1/9(1/9 - x²)] dx

=   ∫ dx/9 [ (1/3)² - x²)

=   (1/9) ∫ dx/ [ (1/3)² - x²)

Formula:

∫ dx/(a² - x²) = (1/2a) log [(a + x)/(a - x)] + c

standard integrals

The previous step exactly matches the above formula. In this we have "1/3" instead of "a" and "x" instead of "x".

=   (1/9) {1/2(1/3) x log [ (1/3 + x)/(1/3 - x) ]  + C}

=   (1/9) {3/2 x log [ (1 + 3x) / 3/(1 - 3x) / 3 ]  + C}

=   (1/9) {3/2 x log [ [(1 + 3x) / 3 ]  [ 3 /(1 - 3x)] ]  + C}

=   (1/3) x 1/2  log [(1 + 3x) /(1 - 3x)] + C

=   (1/6) log [(1 + 3x) /(1 - 3x)] + C

Related pages 