(r2 +)(r2+2)(x2 +3) (2 +4)18.

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Solution

The integral is given as,

I= ∫ ( x 2 +1 )( x 2 +2 ) ( x 2 +3 )( x 2 +4 ) dx

Simplify the given integration.

( x 2 +1 )( x 2 +2 ) ( x 2 +3 )( x 2 +4 ) =1− ( 4 x 2 +10 ) ( x 2 +3 )( x 2 +4 )

Use partial fraction rule.

4 x 2 +10 ( x 2 +3 )( x 2 +4 ) = Ax+B x 2 +3 + Cx+D x 2 +4 4 x 2 +10=( Ax+B )( x 2 +4 )+( Cx+D )( x 2 +3 ) 4 x 2 +10= x 3 ( A+C )+ x 2 ( B+D )+x( 4A+3C )+( 4B+3D )

Equate the coefficients of x 3 , x 2 ,xand constant terms.

A+C=0 B+D=4 4A+3C=0 4B+3D=10

By solving the above equation, we get

A=C=0, B=−2, D=6

Substitute the values and reduce the integral.

I= ∫ dx− ∫ ( 4 x 2 +10 ) ( x 2 +3 )( x 2 +4 ) dx I=x−[ ∫ −2dx ( x 2 +3 ) +6 ∫ dx ( x 2 +4 ) ] I=x−[ −2 ∫ dx x 2 + ( 3 ) 2 +6 ∫ dx ( x 2 + 2 2 ) ]

On integrating, we get

I=x+ 2 3 tan −1 x 3 − 6 2 tan −1 x 2 I=x+ 2 3 tan −1 x 3 −3 tan −1 x 2 +C

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Similar questions

\sum_{r = 1}^{n} \frac{1^{2} + 2^{2} + 3^{2} + . . . . . . . + r^{2}}{r (r + 1)}

r_{1}, r_{2}, r_{1}^{'}

r_{2}^{'}

t_{r} = \frac{1^{2} + 2^{2} + 3^{2} + . . . + r^{2}}{1^{3} + 2^{3} + . . . + r^{3}}

S_{n} = n \sum r = 1 {(- 1)}^{r} . t_{r},

lim n \to \infty S_{n} = \frac{2}{3}

1^{2} + 2^{2} + 3^{2} + . . . + r^{2} = \frac{r (r + 1) (2 r + 1)}{6}

1^{3} + 2^{3} + 3^{3} + . . . + r^{3} = {(\frac{r (r + 1)}{2})}^{2}

Let $c_{1} & c_{2}$ be the centers and $r_{1} & r_{2}$ be the radius of two circles. Then

Cases Conditions

p. 1. $| r_{1} - r_{2} | < c_{1} c_{2} < r_{1} + r_{2}$

q. 2. $| r_{1} - r_{2} | = c_{1} c_{2}$

r. 3. $c_{1} c_{2} < | r_{1} + r_{2} |$

4. $r_{1} + r_{2} < c_{1} r_{2}$

t. 5. $r_{1} + r_{2} = c_{1} r_{2}$

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