Mind-Blowing Calculus Problem in 60 seconds!

The first thing we note is that this quadratic is such that b squared minus 4ac is less than zero, which means that it has no real roots and therefore cannot be factorised easily. However, we can complete the square to write the denominator as follows. Therefore, the integral can be written as x plus 1 plus 1 over x plus 1 squared plus 1. Here, I would suggest to pause the video to think about what the next steps might be. Okay, so we can use the substitution u equals to x plus 1 so that du by dx equals to 1. Therefore, the integral becomes u plus 1 over u squared plus 1 du. Now, to make it easy to solve, we can split this into two integrals as shown here. For the first integral, we can use another substitution v equals to u squared plus 1 so that dv equals to 2u du. Then writing it in terms of v, we get a half integral of dv over v, which is a half natural log of v or a half natural log of modulus of u squared plus 1. And for the second integral, we recognise that it is simply the tan inverse of v. Finally, replacing u with x plus 1, we get the answer, a half natural log of modulus of x plus 1 squared plus 1 plus tan inverse of x plus 1 plus c. A like really helps. Thank you.