Question

int x² √1 + x³ dx find it ​

Answer 1

Answer:

$\large \bold\red{\frac{2}{9} {(1 + {x}^{3} )}^{ \frac{3}{2} } + c}$

Step-by-step explanation:

We have to simplify,

$\displaystyle\int {x}^{2} \sqrt{1 + {x}^{3} } \: dx$

Let's assume that,

$1 + {x}^{3} = t$

Differentiating both the sides,

We get,

$= > 3 {x}^{2} dx = dt \\ \\ = > {x}^{2} dx = \frac{dt}{3}$

Now,

Substituting the values,

We get,

$= \frac{1}{3} \displaystyle\int \sqrt{t} \: dt \\ \\ = \frac{1}{3} \displaystyle\int {t}^{ \frac{1}{2} } dt$

But,

We know that,

• $\displaystyle\int {x}^{n} = \frac{ {x}^{n + 1} }{n + 1}$

Therefore,

We get,

$= \frac{1}{3} \times \frac{ {t}^{ \frac{3}{2} } }{ \frac{3}{2} } + c \\ \\ = \frac{1}{3} \times \frac{2}{3} {t}^{ \frac{3}{2} } + c \\ \\ = \frac{2}{9} {t}^{ \frac{3}{2} } + c$

Substituting the value of t,

We get,

$= \large \bold{\frac{2}{9} {(1 + {x}^{3} )}^{ \frac{3}{2} } + c}$

Where,c is any Constant.

Answer 2

Step-by-step explanation:

We have,

$\displaystyle \sf \int x² \sqrt{1 + x³} dx$

Let $\sf t= 1+ x^3$

differentiating both sides w.r.t x,

$\sf \: \dfrac{dt}{dx} = 3x^2$

$\sf dx = \dfrac{dt}{3x^2}$

$\displaystyle \sf \int x² \sqrt{1 + x³} dx$

Multiply and divide by 3,

$\displaystyle \sf \int \dfrac{3x² \sqrt{1 + x³} }{3}dx$

$\displaystyle \sf \dfrac{1}{3}\int 3x² \sqrt{1 + x^3} dx$

Put the value of dx,

$\displaystyle \sf \dfrac{1}{3}\int 3x² \sqrt{1 + x^3} \dfrac{dt}{3x^2}$

$\displaystyle \sf \dfrac{1}{3}\int \cancel{ 3x²} \sqrt{1 + x^3} \dfrac{dt}{\cancel{3x^2}}$

Put the value of 1+x³ as t

$\displaystyle \sf \dfrac{1}{3}\int \sqrt{t} dt$

$\sf \dfrac{1}{3} × \dfrac{2}{3} t ^{\frac{3}{2}}$

$\sf \dfrac{2}{9} t^{\frac{3}{2}}$

Put the value t as 1 +x³

$\sf \dfrac{2}{9} (1+x^3)^{\frac{3}{2}}$

Add arbitrary constant (C),

$\sf \dfrac{2}{9} (1+x^3)^{\frac{3}{2}} +C$