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Pattern - 39 of Indefinite Integration

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Pattern – 39 :

\int\frac{ax+b}{\left(px^2+qx+r\right)\ \sqrt{mx+n\ }}dx or

\int\frac{1}{\left(px^2+qx+r\right)\ \sqrt{mx+n}}dx , then put mx+n=t^2

 

1) Evaluate : \ \int\frac{x+2}{\left(x^2+3x+3\right)\sqrt{x+1\ }}dx

Solution : Let,

I= \ \int\frac{x+2\ }{\left(x^2+3x+3\right)\sqrt{x+1\ }}dx

Let x + 1 = t^2 & dx = 2t dt

∴  x = t^2 – 1

Now,

x^2+3x+3=\ \left(t^2-1\right)^2+3\ \left(t^2-1\right)+3

=t^4-2t^2+1+3t^2 – 3 + 3

= t^4+t^2 + 1

\therefore\ \ I=\ \int\frac{\left(t^2-1+2\right)}{\left(t^4+t^2+1\right)\ \sqrt{t^2}}\times 2t\ dt

=2\int\frac{t^2+1}{t^4+t^2+1}dt

=2\int\frac{1+\frac{1}{t^2}}{t^2+1+\ \frac{1}{t^2}}\ dt

=2\ \int\frac{\left(1+\ \frac{1}{t^2}\right)}{\left(t-\frac{1}{t}\right)^2+3}\ dt

Let, \ t-\frac{1}{t}=z\ \ \ \therefore\ \ \left(1+\frac{1}{t^2}\right)dt=dz

=\ 2\int\frac{1}{z^2+\ \left(\sqrt3\right)^2}\ dz

=2\times\frac{1}{\sqrt3}\tan^{-1}{\frac{2}{\sqrt3}}+c

=\ \frac{2}{\sqrt3}\tan^{-1}{\frac{t-\frac{1}{t}}{\sqrt3}}+c

=\ \frac{2}{\sqrt3}\tan^{-1}{\frac{t^2-1}{\sqrt3\ t}}+c

=\ \frac{2}{\sqrt3}\tan^{-1}{\frac{x^2}{\sqrt{3\left(x+1\right)}}}+c

 

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