Mathematics

# Solve :$\displaystyle \int \dfrac{\sqrt{1-x^3}}{x^2\sqrt{x}}dx.$ ?

##### SOLUTION
$\quad \int { \dfrac { \sqrt { 1-{ x }^{ 3 } } }{ { x }^{ 2 }\sqrt { x } } dx } \\ =\left( \dfrac { 1 }{ x\sqrt { x } } -\dfrac { 5 }{ 3{ x }^{ \dfrac { 3 }{ 2 } } } \right) \sqrt { 1-{ x }^{ 3 } } -\int { \sqrt { \dfrac { x }{ 1-{ x }^{ 3 } } } } dx\quad \quad \quad \quad \left[ applying\quad integration\quad by\quad parts \right] \\ =\left( \dfrac { 1 }{ x\sqrt { x } } -\dfrac { 5 }{ 3{ x }^{ \dfrac { 3 }{ 2 } } } \right) \sqrt { 1-{ x }^{ 3 } } -\dfrac { 2 }{ 3 } \sin^{-1} { \left( { x }^{ \dfrac { 3 }{ 2 } } \right) } +C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Subjective Medium
Solve  it
$2I=\int _{ O }^{ Q }{ dx }$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate $\displaystyle\int_{0}^{2} 3x+2\ dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
If $\displaystyle I = \int_{0}^{2\pi} e^{x} \cos \left( \frac{x}{2}+\frac{\pi }{4}\right) dx$ then $I$ equals
• A. $\displaystyle \frac{-3}{5} (e^{2\pi}-1)$
• B. $\displaystyle \frac{-3\sqrt{2}} {5} (e^{2\pi}-1)$
• C. $0$
• D. $\displaystyle \frac{-3\sqrt{2}}{5} (e^{2\pi}+1)$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate: $\displaystyle \int { { a }^{ 3\log _{ a }{ x } } } dx$.

Given that for each $\displaystyle a \in (0, 1), \lim_{h \rightarrow 0^+} \int_h^{1-h} t^{-a} (1 -t)^{a-1}dt$ exists. Let this limit be $g(a)$
In addition, it is given that the function $g(a)$ is differentiable on $(0, 1)$