some integrals

for 3rd one substitute \sqrt{x}=z

so we get integral of form 2\int z^2(1+z^4)^{-7/4}dz

now substitute t=\left(1+z^{-4} \right)^{\frac{1}{4}}

this is analogous to integral of form \int x^m(a+bx^n)^pdx

if \left(\frac{m+1}{n}+P \right) is an integer and P is a fraction we put a+bx^n=t^kx^n wer k is deno of fraction P

2nd one
-\int \frac{x^3.dx}{-x^7\left(1+\frac{1}{x^6} \right)^{\frac{1}{2}}} \\ \text{apply parts}\\ -I=2x^3\sqrt{1+\frac{1}{x^6}}-6\underline{\int \left(\frac{\sqrt{1+x^6}}{x} \right)}+C\\ \\ \text{evaluating underlined part}\\ {1+x^6}=z^2\Rightarrow \frac{2z.dz}{6x^5}=dx\\ \int \frac{2z^2.dz}{6(z^2-1)}\text{this can be evaluated easily }

for second one

this integral \int x^{-1}(x^6+1)^{-\frac{1}{2}}dx

is analogous to integral of form \int x^m(a+bx^n)^pdx

if \frac{m+1}{n} is an integer and P is a fraction put (a+bx^n)=t^k wer k is teh deno of fraction P

so substitute (1+x^6)^{\frac{1}{2}}=t

Omkar....these question seems to be from Brilliant YG files..
in the second one substitute x^{3} = t
3x^{2}dx = dt

then u will get \frac{1}{3}\int \frac{dt}{t\sqrt{t^{2}+1}}

then substitue t^{2}+1=z^{2}

tdt = zdz

then u will get

1/3*\int \frac{dz}{z^{2}-1}

which can be easily integrated now using the formula

2. ∫dx/(x⁴√(1+x^-6)) or ∫x^-4dx/√(1+(x^-3)²) now take x^-3=z .....

Q3> put \; x=t^{2}

\Rightarrow\; I=\int \frac{2t^{2}dt}{(1+t^{4})^{\frac{7}{4}}}

take t⁴ common from deno from d bracket

\Rightarrow\; I=\int \frac{2t^{2}dt}{t^{7}(\frac{1}{t^{4}}+1)^{\frac{7}{4}}}

\Rightarrow\; I=\int \frac{2t^{-5}dt}{(\frac{1}{t^{4}}+1)^{\frac{7}{4}}}

now take (\frac{1}{t^{4}}+1)=m, -4t^{-5}dt=dm

\Rightarrow\; I=-\int \frac{dm}{2m^{\frac{7}{4}}}