Blog

Parin Sir > Mathematical Proofs > Indefinite Integration > Pattern - 40 of Indefinite Integration > Pattern - 40 of Indefinite Integration

Pattern - 40 of Indefinite Integration

No Comments

⋆ Pattern – 40

\int\frac{1}{\left(px+q\ \right)\sqrt{ax^2+bx+c}}\ dx, then put px+q=\frac{1}{t}

1) Evaluate : \ \int\frac{1}{\left(2x+3\right)\sqrt{x^2+5x-7}}\ dx

Solution : Let,

I= \ \int\frac{1}{\left(2x+3\right)\sqrt{x^2+5x-7}}\ dx

Let, 2x + 3 = \frac{1}{y}    &

∴  2x = \frac{1}{y} – 3

= \frac{1-3y}{y}

∴  x = \frac{1-3y\ }{2y}

 2dx = \frac{-1}{y^2}\ dy

\therefore\ \ dx=\ \frac{-1}{2y^2}dy

x^2 + 5x – 7

=\ \left(\frac{1-3y}{2y}\right)^2+5\left(\frac{1-3y}{2y}\right)-7

=\ \frac{1-6y+9y^2}{4y^2}+\ \frac{5-15y}{2y}-7

=\ \frac{1-6y+9y^2+10y-30y^2-28y^2}{4\ y^2}

=\ \frac{-49\ y^2+4y+1}{4y^2}

Here, I=\ \int\frac{1}{\left(2x+3\right)\sqrt{x^2+5x-7}}\ dx---\left(1\right)

= \frac{-1\ }{2}\int\frac{1}{\frac{1}{y}\ \sqrt{\frac{-49\ y^2+4y+1\ }{4y^2}}}\times\frac{1}{y^2}\ dy

=\ \frac{-1}{2}\ \int\frac{y\times2y}{\sqrt{-49\ y^2+4y+1}}\times\frac{1}{y^2}\ dy

=\ -\int\frac{1}{\sqrt{1-\left(49y^2-4y\right)}}\ dy

=\ -\int\frac{1}{\sqrt{1-\left(49y^2-4y+\frac{4}{49}\right)+\ \frac{4}{49}}}\ dy

=\ -\ \int\frac{1}{\sqrt{\left(\frac{\sqrt{53}}{7}\right)^2-\ \left(7y-\frac{2}{7}\right)^2}}\ dy

=\ \frac{-1}{7}\sin^{-1}{\left(\frac{7y-\frac{2}{7}}{\frac{\sqrt{53}}{7}}\right)}+c

=\ \frac{-1}{7}\sin^{-1}{\left(\frac{49\ y-2}{\sqrt{53}}\right)}+c

 

Click here to see next pattern

Share
math ganit integration pattern indefinite sabiti

Leave a Reply