$\int \dfrac {\sqrt{x+1}} {x^{7/2}} dx$ without using trigonometry? $$\int \dfrac {\sqrt{x+1}} {x^{7/2}} dx$$
Is there any way to find the answer without using trigonometry, like this? 
Hint by Parth Thakkat:
$$\int \dfrac {\sqrt{x+1}} {x^{7/2}} dx$$
$$ = \int \dfrac {\sqrt{x+1}} {x^{1/2}} \cdot \dfrac{dx} {x^{3}}$$
$$ = \int \sqrt{1+\dfrac 1 x} \cdot \dfrac{dx} {x^{3}}$$
Take $t^2 = 1 + \dfrac 1 x$
and note that $ \dfrac 1 x = t^2 - 1$
More Hint:
$t^2 = 1 + \dfrac 1 x$
$\implies 2tdt = -\dfrac 1 {x^2}dx$
Next
$$ = \int t dx/x^3$$
Since $dx = -2tx^2 dt$
$$= \int t(-2t)(t^2-1) dt$$
$$ = -2\int t^4-t^2 dt$$
$$ = -2(\dfrac{t^5} {5} - \dfrac{t^3}{3} + C$$ 
C = constant
$$ = -\dfrac{2}{5}(1+\dfrac{1}{x})^\dfrac{5}{2}+\dfrac{2}{3}(1+\dfrac{1}{x})^\dfrac{3}{2} + C$$
Thanks alot :)

Hint:
$$\int \dfrac {\sqrt{x+1}} {x^{7/2}} dx$$
$$ = \int \dfrac {\sqrt{x+1}} {x^{1/2}} \cdot \dfrac{dx} {x^{3}}$$
$$ = \int \sqrt{1+\dfrac 1 x} \cdot \dfrac{dx} {x^{3}}$$
Take $t^2 = 1 + \dfrac 1 x$
and note that $ \dfrac 1 x = t^2 - 1$
More Hint:
$t^2 = 1 + \dfrac 1 x$
$\implies 2tdt = -\dfrac 1 {x^2}dx$
$\implies 2t(1-t^2)dt = \dfrac 1 {x^3}dx$
