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tt4+t25dt is equal to 

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a
121log2t2+1+212t2+1+21+c
b
121log2t2−1+212t2+1+21+c
c
1221log2t2−1+212t2+1+21
d
1221log2t2+1−212t2+1+21+c

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detailed solution

Correct option is D

∫tt4+t2−5dtt2=z2tdt=dz∫tt22+t2−5dt=∫dzz2+z−5=12∫dzz2+2.z12+122−5−122 Use ∫1x2−a2dx=12alog⁡x−ax+a+c=12∫dzz+122−214=12∫dzz+122−2122=1212212log⁡z+12−212z+12+212+c=1221log⁡2z+1−212z+1+21+c


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