Question

integral dx(x^1/2+x^1/3)​

Answer 1

Answer:-

Before solving, remember that,

1. $\displaystyle \int \sf \small(f(x) \pm g(x))dx = \int f(x)dx \pm \int g(x)dx$
2. $\displaystyle \int\sf {x}^{n} dx = \frac{ \large{x}^{n + 1} }{n + 1}$

Now, we will solve,

$\displaystyle \int \sf \small \big( \sqrt{x} + \sqrt[3]{x} )dx$

$\displaystyle \sf = \int \big( \sqrt{x} \big )dx + \int \big( \sqrt[3]{x} \big)dx$

$\sf = \large\frac{ {x}^{ \frac{1}{2} + 1} }{ \frac{1}{2} + 1 } + \frac{ {x}^{ \frac{1}{3} + 1 } }{ \frac{1}{3} + 1}$

$\sf = \large \frac{x \sqrt{x} }{ \frac{3}{2} } + \frac{x \sqrt[3]{x} }{ \frac{4}{3} }$

Now, add the constant of integration,

$\sf = \frac{2x \sqrt{x} }{3} + \frac{3x \sqrt[3]{x} }{4} + C, C \in \R$

Answer 2

Answer:

Before solving, remember that,

\displaystyle \int \sf \small(f(x) \pm g(x))dx = \int f(x)dx \pm \int g(x)dx∫(f(x)±g(x))dx=∫f(x)dx±∫g(x)dx

\displaystyle \int\sf {x}^{n} dx = \frac{ \large{x}^{n + 1} }{n + 1}∫x

n

dx=

n+1

x

n+1

Now, we will solve,

\displaystyle \int \sf \small \big( \sqrt{x} + \sqrt[3]{x} )dx∫(

x

+

3

x

)dx

\displaystyle \sf = \int \big( \sqrt{x} )dx + \int \big( \sqrt[3]{x} \big)dx=∫(

x

)dx+∫(

3

x

)dx

\sf = \large\frac{ {x}^{ \frac{1}{2} + 1} }{ \frac{1}{2} + 1 } + \frac{ {x}^{ \frac{1}{3} + 1 } }{ \frac{1}{3} + 1}=

2

1

+1

x

2

1

+1

+

3

1

+1

x

3

1

+1

\sf = \large \frac{x \sqrt{x} }{ \frac{3}{2} } + \frac{x \sqrt[3]{x} }{ \frac{4}{3} }=

2

3

x

x

+

3

4

x

3

x

Now, add the constant of integration,

\sf = \frac{2x \sqrt{x} }{3} + \frac{3x \sqrt[3]{x} }{4} + C, C \in \R=

3

2x

x

+

4

3x

3

x

+C,C∈R