$\int \dfrac {\sqrt{x+1}} {x^{7/2}} dx$ sin utilizar la trigonometría?

Question

$$\int \dfrac {\sqrt{x+1}} {x^{7/2}} dx$$

¿Hay alguna manera de encontrar la respuesta sin usar la trigonometría, como este ?

Pista de Parth Thakkat :

$$\int \dfrac {\sqrt{x+1}} {x^{7/2}} dx$$

$$ = \int \dfrac {\sqrt{x+1}} {x^{1/2}} \cdot \dfrac{dx} {x^{3}}$$

$$ = \int \sqrt{1+\dfrac 1 x} \cdot \dfrac{dx} {x^{3}}$$

Toma $t^2 = 1 + \dfrac 1 x$

y nota que $ \dfrac 1 x = t^2 - 1$

Más pistas :

$t^2 = 1 + \dfrac 1 x$

$\implies 2tdt = -\dfrac 1 {x^2}dx$

Siguiente

$$ = \int t dx/x^3$$

Desde $dx = -2tx^2 dt$

$$= \int t(-2t)(t^2-1) dt$$ $$ = -2\int t^4-t^2 dt$$ $$ = -2(\dfrac{t^5} {5} - \dfrac{t^3}{3} + C$$ C = constante

$$ = -\dfrac{2}{5}(1+\dfrac{1}{x})^\dfrac{5}{2}+\dfrac{2}{3}(1+\dfrac{1}{x})^\dfrac{3}{2} + C$$

Muchas gracias :)

Answer 1

Sugerencia :

$$\int \dfrac {\sqrt{x+1}} {x^{7/2}} dx$$

$$ = \int \dfrac {\sqrt{x+1}} {x^{1/2}} \cdot \dfrac{dx} {x^{3}}$$

$$ = \int \sqrt{1+\dfrac 1 x} \cdot \dfrac{dx} {x^{3}}$$

Toma $t^2 = 1 + \dfrac 1 x$

y nota que $ \dfrac 1 x = t^2 - 1$

Más pistas :

$t^2 = 1 + \dfrac 1 x$

$\implies 2tdt = -\dfrac 1 {x^2}dx$

$\implies 2t(1-t^2)dt = \dfrac 1 {x^3}dx$

Answer 2

$$\int \sqrt{x+1}/x^{7/2} dx=\int\frac{1}{x^3} \sqrt{1+\frac{1}{x}} dx$$ $1+\frac{1}{x}=t$ entonces $\frac{1}{x^2}dx=-dt $ y $\frac{1}{x}={t-1}$ $$\int\frac{1}{x^3} \sqrt{1+\frac{1}{x}} dx=\int\frac{1}{x} \sqrt{1+\frac{1}{x}} \frac{dx}{x^2}=-\int(t-1)\sqrt {t}dt=\int (t^{1/2}-t^{3/2})dt$$