(x+1)√x+3 .dx integral
Answers
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:
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