We are given that,I=∫(x2+1)(x2−5x+6)dx
=(x2−5x+6+5x−5)x2−5x+6dx
I=∫(x2−5x+6x2−5x+6)dx+∫(5x−5x2−5x+6)dx
I=∫dx+∫5x−5x2−5x+6dx
Let I1=∫dx=x+C1,C1∈R
& I2=∫5x−5x2−5x+6dx.
I2=∫52(2x−5)+152(x2−5x+6)dx
So, I2=52∫(2x−5)(x2−5x+6)dx+152∫dxx2−5x+6
Let I3=52∫(2x−5)(x2−5x+6=52ln|x2−5x+6|+C3,C3∈R
& I4=152∫dxx2−5x+6
I4=152∫dx(x−52)2−(√32)2
So, I4=152⎡⎢
⎢
⎢
⎢
⎢
⎢⎣12(√152)loge∣∣
∣
∣
∣∣x−52−√132x−52+√132∣∣
∣
∣
∣∣+C4⎤⎥
⎥
⎥
⎥
⎥
⎥⎦ C4∈R
So, I=I1+I2
where I2=I3+I4
I2=52ln|x2−5x+6|+C3+152(√213)loge∣∣∣2x−5−√132x−5+√13∣∣∣+C4
I=x+C1+52ln|x2−5x+6|+15√2×1√13loge∣∣∣2x−5−√132x−5+√13∣∣∣+C2
Where C2=C3+C3
I=x+52ln|x2−5x+6|+15√26loge∣∣∣2x−5√132x−5+√13∣∣∣+C
where C∈R & C=C1+C2
∫x2+1x2−5x+6dx=x+52ln|x2−5x+6|+15√26loge∣∣∣2x−5−√132x−5+√13∣∣∣+C