1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# If f(x)=x+22x+3, then ∫(f(x)x2)1/2dx=1√2g(1+√2f(x)1−√2f(x))−√23h(√3f(x)+√2√3f(x)−√2)+c, where

A
g(x)=tan1x,h(x)=log|x|
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
g(x)=log|x|,h(x)=tan1x
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
g(x)=h(x)=tan1x
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
g(x)=log|x|,h(x)=log|x|
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is D g(x)=tan−1x,h(x)=log|x|I=∫√x+2x√2x+3dxSubstitute u=√2x+3⇒dx=√2x+3du, use 1x=2u2−3,x=u2−32:=∫√2√u2+1u2−3duPerform u=tan(v)⇒v=tan−1u,du=sec2vdv:∫sec2(v)√tan2(v)+1tan2(v)−3dvSimplify using tan2(v)+1=sec2(v):=∫sec3(v)tan2(v)−3dv=∫cos(v)(−1(sin2(v)−1)(4sin2(v)−3))dvSubstitute w=sin(v)⇒dv=1cos(v)dw:=−∫1(w2−1)(4w2−3)dwFactor the denominator and perform partial fraction decomposition:=−∫(−44w2−3−12(w+1)+12(w−1))dw=−(−4∫14w2−3dw−12∫1w+1dw+12∫1w−1dw) ........(i)Now, ∫14w2−3dw=∫1(2w−√3)(2w+√3)dwPerform partial fraction decomposition:=∫⎛⎜ ⎜⎝12.√3(2w−√3)−12.√3(2w+√3)⎞⎟ ⎟⎠dw=12.√3∫12w−√3dw−12.√3∫12w+√3dw=ln(2w−√3)4.√3−ln(2w+√3)4.√3So, −4∫14w2−3dw−12∫1w+1dw+12∫1w−1dw→=−ln(2w−√3)√3+ln(2w+√3)√3−ln(w+1)2+ln(w−1)2Plug in solved integrals:−∫1(w2−1)(4w2−3)dw=ln(2w−√3)√3−ln(2w+√3)√3+ln(w+1)2−ln(w−1)2Undo substitution w=sin(v):=ln(2sin(v)−√3)√3−ln(2sin(v)+√3)√3+ln(sin(v)+1)2−ln(sin(v)−1)2Undo substitution v=tan−1(u), use sin(tan−1(u))=u√u2+1:=ln(2u√u2+1−√3)√3−ln(2u√u2+1+√3)√3+ln(u√u2+1+1)2−ln(u√u2+1−1)2Plug in solved integrals:√2∫√u2+1u2−3du=√23ln(2u√u2+1−√3)−√23ln(2u√u2+1+√3)+ln(u√u2+1+1)√2−ln(u√u2+1−1)√2Undo substitution u=√2x+3⇒uu2+1=√(2x+3)(2x+4)=1√2f(x)=√23ln(√2f(x)−√3)−√23ln(√2f(x)+√3)+ln(1√2f(x)+1)√2−ln(1√2f(x)−1)√2Now, taking LCM inside log function and then using lnA−lnB=ln(AB):we get, I=1√2ln(1+√2f(x)1−√2f(x))−√23ln(√3f(x)+√2√3f(x)−√2)+cSo, g(x)=ln|x|,h(x)=ln|x|Hence, option (D) is correct.

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Property 7
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program