The given equation can be written as

⇒

First integration be I1 and second be I2.

⇒ For I1

Add and subtract 2 from the numerator

⇒

⇒

⇒

⇒ x - 2ln|x + 2| + c1

∴ I1 = x - 2ln|x + 2| + c1

For I2

⇒

Assume x + 1 = t

dt = dx

⇒

Substitute u = √t

dt = 2√t.du

t = u²

⇒

Add and subtract 1 in the above equation:

⇒

⇒

⇒

⇒ 2u - tan ^{- 1}(u) + c2

But u = √t

∴ 2√t - tan ^{- 1}(√t) + c2

Also t = x + 1

∴ 2√(x + 1) - tan ^{- 1}(x + 1) + c2

I = I1 + I2

∴ I = x - 2ln|x + 2| + c1 + 2√(x + 1) - tan ^{- 1}(x + 1) + c2

I = x - 2ln|x + 2| + 2√(x + 1) - tan ^{- 1}(x + 1) + c.