1kya hai yaar [5]
3 Answers
62x+2=t
then this becomes t^5(t+1)^2
which you can write as \\\frac{1}{t^5(t+1)^2} \\\frac{t^2+2t+1-t^2-2t}{t^5(t+1)^2} \\\frac{(t+1)^2-t^2-2t}{t^5(t+1)^2} \\\frac{1}{t^5}+\frac{1}{t^3(t+1)^2}-\frac{2}{t^4(t+1)^2}
now you can keep moving to get the answer...
1708Thanks Nishant Sir.
My solution::
\mathbf{\int\frac{1}{(x+2)^5(x+3)^3}dx}$\\\\ Now Put $\mathbf{x+2=t\Leftrightarrow dx=dt}\\\\ \mathbf{\int\frac{dt}{t^5(t+1)^3}dt}$\\\\ Now Put $\mathbf{t=\frac{1}{a}\Leftrightarrow dt=-\frac{1}{a^2}da}$\\\\ $\mathbf{-\int\frac{a^6}{(1+a)^3}da}$\\\\ Now Again Put $\mathbf{1+a=p\Leftrightarrow da=dp}$\\\\ So $\mathbf{-\int\frac{(1-p)^6}{p^3}dp}$\\\\ $\mathbf{-\int\frac{p^6-6p^5+15p^4-20p^3+15p^2-6p+1}{p^3}dp}$\\\\ $\mathbf{-\int\left(p^3-6p^2+15p-20+\frac{20}{p}-\frac{6}{p^2}+\frac{1}{p^3}\right)dp}$\\\\ $\mathbf{-\left(\frac{p^4}{4}-2p^3+\frac{15p^2}{2}-20p+15ln(p)+\frac{6}{p}-\frac{1}{2p^2}\right)+C}$\\\\ Now Put value of $\mathbf{p}$ and then $\mathbf{a}$ and Then $\mathbf{t}$.