Evaluate the following integrals:

The given equation can be written as
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First integration be I1 and second be I2.
⇒ For I1
Add and subtract 2 from the numerator
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⇒ x - 2ln|x + 2| + c1
∴ I1 = x - 2ln|x + 2| + c1
For I2
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Assume x + 1 = t
dt = dx
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Substitute u = √t
dt = 2√t.du
t = u2
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Add and subtract 1 in the above equation:
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⇒ 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.
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