1) Integral {(x^3)-4x}/{(x^2)+1}^2.what Is The Solution?
2) Integral {x^(1/2)}dx/{1+ X^(1/4)}.what Is The Solution?
Please help me out.i require detailed solutions.
thaking you.
1. $\displaystyle \displaystyle \int \frac{x^3-4x}{(x^2+1)^2}\;{dx}$ $\displaystyle \displaystyle = \int \frac{x(x^2-4)}{(x^2+1)^2}\;{dx}$
Let $\displaystyle \displaystyle t = x^2+1 \Rightarrow dx = \frac{1}{2x}\;{dt}$
Then also $\displaystyle x^2-4 = t-5 $
We have $\displaystyle \displaystyle \int \frac{x^3-4x}{(x^2+1)^2}\;{dx}$ $\displaystyle \displaystyle = \frac{1}{2}\int \frac{x(t-5)}{xt^2}\;{dt} = \frac{1}{2}\int \left(\frac{1}{t}-\frac{5}{t^2}\right)\;{dt} = \ln{t}+\frac{5}{t}+k$
Hence $\displaystyle \displaystyle \int \frac{x^3-4x}{(x^2+1)^2}\;{dx} = \frac{(x^2+1)\ln(x^2+1)+5}{2(x^2+1)}+k$
Thank you very much to you both for replying.I tried the first using partial fractions and substitution,but got stuck and the second one i still have no idea how to go forward.
Thank you TheCoffeeMachine.If it is is'nt too much for you could you please help me with the second one too.
Just in case a picture helps...
In fact, a choice of two - either...
... or...
... where (key in spoiler) ...
Spoiler:
Either way, work anti-clockwise, and find G then F then I.
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