Can someone help me with this question..
For the integral $\displaystyle \int{\frac{1}{( x^2 + 1 )^n}} dx $ prove the reduction formula :
$\displaystyle I_{n} = \frac{1}{n-2} . \frac{x}{(x^2+1)^{n-1}} + \frac{2n - 3}{2n-2}I_{n-1}$
Thanks
Can someone help me with this question..
For the integral $\displaystyle \int{\frac{1}{( x^2 + 1 )^n}} dx $ prove the reduction formula :
$\displaystyle I_{n} = \frac{1}{n-2} . \frac{x}{(x^2+1)^{n-1}} + \frac{2n - 3}{2n-2}I_{n-1}$
Thanks
How to find the integral of
a) $\displaystyle \int{\frac{3}{\sqrt{6 + 4x - x^2}}} $
and
b) $\displaystyle \int{\frac{3x^2}{\sqrt{x^2 - 9}}} $
for b) The answer in the ans sheet is this
$\displaystyle \frac{3x \sqrt{x^2 - 9}}{2} + \frac{27}{2} ln (x + \sqrt{x^2 -9}) + C $
What I've found is this :
$\displaystyle \frac{27}{3} ln (\frac{1}{3} + \sqrt{x^2 -9}) $
and
c) $\displaystyle \int{x^3 ln (ax)} $
Ans : $\displaystyle -\frac{x^4}{10}+ \frac{x^4 ln (ax)}{4} $
My ans : $\displaystyle -\frac{x^4}{16}+ \frac{x^4 ln (ax)}{4} $