Advertisements
Advertisements
Question
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
Sum
Advertisements
Solution
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}}dx\]
\[Let, 1 + \sqrt{x} = t\]
\[ \Rightarrow \frac{1}{2\sqrt{x}} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{dx}{\sqrt{x}} = 2dt\]
\[Now, \int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}}dx \]
\[ = 2\int t^2 dt\]
\[ = \frac{2}{3} t^3 + C\]
\[ = \frac{2}{3} \left( 1 + \sqrt{x} \right)^3 + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\frac{x^5 + x^{- 2} + 2}{x^2} dx\]
\[\int\frac{1}{1 - \sin x} dx\]
\[\int\frac{2x - 1}{\left( x - 1 \right)^2} dx\]
\[\int\frac{x + 1}{\sqrt{2x + 3}} dx\]
\[\int\frac{2 - 3x}{\sqrt{1 + 3x}} dx\]
\[\int\frac{\text{sin} \left( x - a \right)}{\text{sin}\left( x - b \right)} dx\]
\[\int\frac{\left( x + 1 \right) e^x}{\cos^2 \left( x e^x \right)} dx\]
\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} dx\]
\[\int \tan^3 \text{2x sec 2x dx}\]
\[\int \cot^5 \text{ x } {cosec}^4 x\text{ dx }\]
\[\int x \cos^3 x^2 \sin x^2 \text{ dx }\]
\[\int\frac{1}{\sin^3 x \cos^5 x} dx\]
\[\int\frac{3 x^5}{1 + x^{12}} dx\]
\[\int\frac{x^2}{x^6 + a^6} dx\]
\[\int\frac{x}{x^4 - x^2 + 1} dx\]
\[\int\frac{\cos x}{\sqrt{\sin^2 x - 2 \sin x - 3}} dx\]
\[\int\frac{x + 1}{\sqrt{x^2 + 1}} dx\]
\[\int\frac{1}{3 + 2 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{1 - \sin x + \cos x} \text{ dx }\]
\[\int\frac{1}{2 + \sin x + \cos x} \text{ dx }\]
\[\int\frac{1}{\sqrt{3} \sin x + \cos x} dx\]
\[\int\frac{1}{1 - \tan x} \text{ dx }\]
\[\int x e^x \text{ dx }\]
\[\int \tan^{- 1} \left( \sqrt{x} \right) \text{dx }\]
\[\int\frac{e^x}{x}\left\{ \text{ x }\left( \log x \right)^2 + 2 \log x \right\} dx\]
\[\int e^x \cdot \frac{\sqrt{1 - x^2} \sin^{- 1} x + 1}{\sqrt{1 - x^2}} \text{ dx }\]
\[\int\left( 2x + 3 \right) \sqrt{x^2 + 4x + 3} \text{ dx }\]
\[\int\frac{x}{\left( x + 1 \right) \left( x^2 + 1 \right)} dx\]
\[\int\frac{1}{x \left( x^4 - 1 \right)} dx\]
\[\int\frac{x^2}{\left( x - 1 \right) \sqrt{x + 2}}\text{ dx}\]
\[\int\frac{x}{\left( x - 3 \right) \sqrt{x + 1}} \text{ dx}\]
\[\int\frac{1}{\left( x + 1 \right) \sqrt{x^2 + x + 1}} \text{ dx }\]
\[\int\frac{1}{\cos x + \sqrt{3} \sin x} \text{ dx } \] is equal to
If \[\int\frac{1}{5 + 4 \sin x} dx = A \tan^{- 1} \left( B \tan\frac{x}{2} + \frac{4}{3} \right) + C,\] then
\[\int\frac{1}{1 + \tan x} dx =\]
\[\int\frac{x^2}{\left( x - 1 \right)^3} dx\]
\[\int\sqrt{a^2 - x^2}\text{ dx }\]
\[\int x\sqrt{1 + x - x^2}\text{ dx }\]
\[\int\left( 2x + 3 \right) \sqrt{4 x^2 + 5x + 6} \text{ dx}\]
\[\int\frac{e^{m \tan^{- 1} x}}{\left( 1 + x^2 \right)^{3/2}} \text{ dx}\]
