Question

(x+1)√x+3 .dx integral
​

Answer 1

Answer:

Answer:

x = 9/13 or x = 4/13

Step-by-step explanation:

Here, the given expression,

\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}

1−x

x

+

x

1−x

=

6

13

\frac{\sqrt{x}}{\sqrt{1-x}}+\frac{\sqrt{1-x}}{\sqrt{x}}=\frac{13}{6}

1−x

x

+

x

1−x

=

6

13

\frac{x+1-x}{\sqrt{x(1-x)}}=\frac{13}{6}

x(1−x)

x+1−x

=

6

13

\frac{1}{\sqrt{x(1-x)}}=\frac{13}{6}

x(1−x)

1

=

6

13

6=13[\sqrt{x(1-x)}]6=13[

x(1−x)

]

By squaring both sides,

36=169[x(1-x)]36=169[x(1−x)]

36=169x-169x^236=169x−169x

2

169x^2-169x+36=0169x

2

−169x+36=0

By splitting the middle term,

169x^2-117x-52x+36=0169x

2

−117x−52x+36=0

13x(13x-9)-4(13x-9)=013x(13x−9)−4(13x−9)=0

(13x-9)(13x-4)=0(13x−9)(13x−4)=0

\implies 13x - 9 = 0\text{ or }13x - 4 = 0⟹13x−9=0 or 13x−4=0

\implies x = \frac{9}{13}\text{ or }x = \frac{4}{13}⟹x=

13

9

or x=

13

4

Which is the required solution.

Answer 2

Answer:

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