Solve it by simple method $$\int \frac{\mathrm{d}x}{(x^2+1)^4}$$
This is what i did: Let $x=\tan{\alpha}$
After solving we get $\int \cos^6(\alpha)\;\mathrm{d}\alpha$
Again by expanding we get $\frac{1}{2^6}\times(2\cos6\alpha+12\cos4\alpha+30\cos2\alpha+20)+c$ where $x=\tan{\alpha}$
I think this is not good way of solving problem i am not getting answer , Please help with other simple method !
On
One option is going on partial fractions, but it would take for ever to come up with a tractable result.
As an alternative, once you have to integrate $\cos^6 \alpha$, recall:
$$ \int \cos^q \alpha \, \mathrm{d} \alpha = -\frac{\cos^{q+1}(x) \, _2F_1\left(\frac{1}{2},\frac{q+1}{2};\frac{q+3}{2};\cos ^2(x)\right)}{q+1}, \quad q \neq -1,$$ where ${}_2 F_1(a,b;c;z)$ is the hypergeometric function.
I cannot provide any reference (I did the computation with Mathematica), but I'm sure I have seen this somewhere (maybe Abramovitz). I'm afraid nobody would define this as a "simple method", but I think it's a nice result.
Hope you find this useful.
On
Integrating by parts,
$$I_n=\int1\cdot(x^2+1)^ndx$$
$$=(x^2+1)^n\int dx-\int\left[\dfrac{d\left((x^2+1)^n\right)}{dx}\int dx\right]dx$$
$$=x(x^2+1)^n-\int[n(1+x^2)^{n-1}2x^2]dx$$
$$=x(x^2+1)^n-2n\int(1+x^2)^{n-1}(1+x^2-1)dx$$
$$=x(x^2+1)^n-2n[I_n-I_{n-1}]$$
$$\implies(2n+1)I_n=x(x^2+1)^n+2nI_{n-1}$$
I think that the substitution $x=\tan(a)$ is from far away the simplest to use for the calculation of $$I_m=\int \frac{\mathrm{d}x}{(x^2+1)^{m}}=\int \cos^{2(m-1)}(a)\,\mathrm{d}a$$ At this point, the only thing to be done is a power reduction of the cosine which will transform the cosine raised to power into a linear combination of cosines of multiple angles; this is what you did.
The general formulas are quite simple. If $m$ is odd,
$$\cos^m(a)=\frac 1{2^{m-1}}\sum_{k=0}^{\frac{m-1}2}\binom{m}{k} \cos \Big( (m-2 k)\,a\Big)$$ and, if $m$ is even, $$\cos^m(a)=\frac 1{2^{m}}\binom{m}{\frac m2} +\frac 1{2^{m-1}}\sum_{k=0}^{\frac{m}2-1}\binom{m}{k} \cos \Big( (m-2 k)\,a\Big)$$ and the integration does not make any problem.
You could fine similar reduction formulas for $\sin^m(a)$.
If you do not use this substitution, you will face hypergeometric functions (as Dmoreno already mentioned) and $$I_m=x \, _2F_1\left(\frac{1}{2},m;\frac{3}{2};-x^2\right)$$ which can be represented as $$I_m=\frac{P_{2m-3}(x)}{(1+x^2)^{m-1}}-k_m \tan^{-1}(x)$$ where $P_n(x)$ represents a polynomial of degree $n$.