Questions
is equal to
detailed solution
Correct option is A
Let x+2=Addxx2+2x+3+B ⇒ x+2=A(2x+2)+B⇒ x+2=2Ax+(2A+B)On equating the coefficient of x and constant term on both sides, we get 2A=1⇒A=12and 2A+B=2⇒2×12+B=2⇒ B=2-1=1 ⇒ x+2=122x+2+1∴∫x+2x2+2x+3dx=∫12(2x+2)+1x2+2x+3dx =12∫2x+2x2+2x+3dx+∫dxx2+2x+3Let l1=∫2x+2x2+2x+3dx and I2=∫dxx2+2x+3Then,∫x+2x2+2x+3dx=12I1+I2 .....(i)Now, I1=∫2x+2x2+2x+3dxLet x2+2x+3=t⇒(2x+2)dx=dt I2=∫dtt=∫t−1/2dt=t(−1/2)+11/2+1+C=2t+C1 =2x2+2x+3+C1and I2=∫dxx2+2x+3=∫dxx2+2x+(1)2+3−(1)2 =∫dx(x+1)2+(2)2Let x+1=t⇒dx=dtI2=∫dtt2+(2)2=logt+t2+2+C2 ∵∫dxx2+a2=log∣x+x2+a2=logx+1+(x+1)2+2+C2=logx+1+x2+2x+3+C2On putting the values of l1 and l2 in Eq. (i), we get∫x+2x2+2x+3dx=122x2+2x+3+logx+1+x2+2x+3∵12C1+C2=C=x2+2x+3+log∣x+1+x2+2x+3+CSimilar Questions
is equal to
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests