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Question

-5)(7-4)6x719.


Solution

The given integral is,

6x+7 ( x5 )( x4 ) dx

The above integral can be rewritten as,

6x+7 x 2 9x+20 dx

Let,

6x+7=A d dx ( x 2 9x+20 )+B 6x+7=A( 2x9 )+B (1)

Equate the coefficients on both the sides.

2A=6 A=3 9A+B=7 B=34

Substitute the values of A and B in equation (1).

6x+7=3( 2x9 )+34

This satisfies the equation.

Now, substitute the value of 6x+7 from the above expression in equation (1).

6x+7 x 2 9x+20 dx= 3( 2x9 )+34 x 2 9x+20 dx =3 2x9 x 2 9x+20 dx+34 1 x 2 9x+20 dx

Let,

I 1 = 2x9 x 2 9x+20 dx I 2 = 1 x 2 9x+20 dx

The above expression can be rewritten as

6x+7 x 2 9x+20 dx=3 I 1 +34 I 2 (2)

Integrate the term I 1 . Let,

x 2 9x+20=t ( 2x9 )dx=dt

Substitute the value of ( 2x9 )dx in I 1 .

( 2x9 ) x 2 9x+20 dx= dt t 1 2 = t 1 2 + C 1 =2 t + C 1 =2 x 2 9x+20 + C 1 (3)

Rewrite the term x 2 9x+20 as x 2 9x+20+ 81 4 81 4 .

x 2 9x+20= x 2 9x+20+ ( 9 2 ) 2 ( 9 2 ) 2 = ( x 9 2 ) 2 1 4 = ( x 9 2 ) 2 ( 1 2 ) 2

Integrate the term I 2 by substituting the value of x 2 9x+20.

1 x 2 9x+20 dx = 1 ( x 9 2 ) 2 ( 1 2 ) 2 dx =log| ( x 9 2 )+ x 2 9x+20 |+ C 2 (4)

Substitute the values of I 1 and I 2 from equations (3) and (4) in equation (2).

3 I 1 +34 I 2 =3×2 x 2 9x+20 + C 1 +34×log| ( x 9 2 )+ x 2 9x+20 |+ C 2 =6 x 2 9x+20 +34log| ( x 9 2 )+ x 2 9x+20 |+C


Mathematics
Math Part-II - NCERT
Standard XII

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