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  • Integrate the functions in Exercises 1 to 24. 14

Q: Integrate the function 1(x2+1)(x2+4)

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Given, 1(x2+1)(x2+4)

Let I=∫1(x2+1)(x2+4)

Now, Using partial differentiation,

1(x2+1)(x2+4)=Ax+B(x2+1)+Cx+D(x2+4)

⟹1(x2+1)(x2+4)=(Ax+B)(x2+4)+(Cx+D)(x2+1)(x2+1)(x2+4)

⟹1=(Ax+B)(x2+4)+(Cx+D)(x2+1)

⟹1=Ax3+4Ax+Bx2+4B+Cx3+Cx+Dx2+D

⟹(A+C)x3+(B+D)x2+(4A+C)x+(4B+D)=1

Equating the coefficients of x,x2,x3 and constant value,

A+C=0⟹C=−A
B+D=0⟹B=−D
4A+C=0⟹4A=−C⟹4A=A⟹A=0=C
4B+D=1⟹4B−B=1⟹B=13=−D

Putting these values in equation, we have

⇒1(x2+1)(x2+4)=(0)x+13(x2+1)+(0)x+(−13)(x2+4)

1(x2+1)(x2+4)=13(x2+1)+(−13)(x2+4)

⇒∫1(x2+1)(x2+4)dx=13⋅∫1(x2+1)dx−13⋅∫1(x2+4)dx

=13⋅tan−1⁡x−13⋅12tan−1⁡x2+C

⇒I=13tan−1⁡x−16tan−1⁡x2+C

Posted by

HARSH KANKARIA

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