Home page for accesible maths 1 1 Further Integration

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1.18 Simple examples

Example.

To find (1) ∫x2+1x⁢d⁢x\int{{x^{2}+1}\over x}\,dx and (2) ∫x+2x+1⁢d⁢x\int{{x+2}\over{x+1}}\,dx.

Solution.

For (1), we have ∫x2+1x⁢d⁢x=∫(x+1x)⁢d⁢x=x22+log⁡|x|+c\int{{x^{2}+1}\over x}\,dx={\int(x+{1\over x})\,dx={{x^{2}}\over 2}+\log|x|+c} for some constant cc.

For (2), we express x+2x+1{{x+2}\over{x+1}} as (x+1)+1x+1=1+1x+1{{(x+1)+1}\over{x+1}}=1+{{1}\over{x+1}}.

Hence ∫x+2x+1⁢d⁢x=\int{{x+2}\over{x+1}}dx= ∫(1+1x+1)⁢d⁢x=x+log⁡|x+1|+c\int\left(1+\frac{1}{x+1}\right)\,dx=x+\log|x+1|+c for some constant cc.