Advertisements
Advertisements
Question
\[\int\frac{1}{\left( 7x - 5 \right)^3} + \frac{1}{\sqrt{5x - 4}} dx\]
Sum
Advertisements
Solution
\[\int\left[ \frac{1}{\left( 7x - 5 \right)^3} + \frac{1}{\sqrt{5x - 4}} \right]dx\]
\[ = \int\left[ \left( 7x - 5 \right)^{- 3} + \left( 5x - 4 \right)^{- \frac{1}{2}} \right]dx\]
\[ = \frac{\left( 7x - 5 \right)^{- 3 + 1}}{7\left( - 3 + 1 \right)} + \frac{\left( 5x - 4 \right)^{- \frac{1}{2} + 1}}{5\left( - \frac{1}{2} + 1 \right)} + C\]
\[ = \frac{\left( 7x - 5 \right)^{- 2}}{- 14} + \frac{2}{5} \left( 5x - 4 \right)^\frac{1}{2} + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
\[\int\frac{1 - \cos 2x}{1 + \cos 2x} dx\]
\[\int \left( a \tan x + b \cot x \right)^2 dx\]
\[\int\frac{2x - 1}{\left( x - 1 \right)^2} dx\]
\[\int\frac{3x + 5}{\sqrt{7x + 9}} dx\]
` ∫ sin x \sqrt (1-cos 2x) dx `
\[\int\frac{\sec^2 x}{\tan x + 2} dx\]
\[\int\frac{1 + \cot x}{x + \log \sin x} dx\]
\[\int\frac{\sin \left( \tan^{- 1} x \right)}{1 + x^2} dx\]
\[\int\frac{x + \sqrt{x + 1}}{x + 2} dx\]
\[\int\sqrt {e^x- 1} \text{dx}\]
\[\int \sin^3 x \cos^5 x \text{ dx }\]
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 - 1}} dx\]
\[\int\frac{\left( 3\sin x - 2 \right)\cos x}{13 - \cos^2 x - 7\sin x}dx\]
\[\int\frac{x^2 + x + 1}{x^2 - x + 1} \text{ dx }\]
\[\int\frac{2x + 5}{\sqrt{x^2 + 2x + 5}} dx\]
\[\int\frac{\cos x}{\cos 3x} \text{ dx }\]
\[\int\frac{1}{1 - \sin x + \cos x} \text{ dx }\]
\[\int\frac{1}{3 + 2 \sin x + \cos x} \text{ dx }\]
\[\int\frac{1}{p + q \tan x} \text{ dx }\]
\[\int\frac{\log x}{x^n}\text{ dx }\]
\[\int \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\frac{\left( x \tan^{- 1} x \right)}{\left( 1 + x^2 \right)^{3/2}} \text{ dx }\]
\[\int\left( \tan^{- 1} x^2 \right) x\ dx\]
\[\int\frac{x^3 \sin^{- 1} x^2}{\sqrt{1 - x^4}} \text{ dx }\]
\[\int e^x \left( \frac{1}{x^2} - \frac{2}{x^3} \right) dx\]
\[\int\frac{2x - 3}{\left( x^2 - 1 \right) \left( 2x + 3 \right)} dx\]
\[\int\frac{2x}{\left( x^2 + 1 \right) \left( x^2 + 3 \right)} dx\]
\[\int\frac{dx}{\left( x^2 + 1 \right) \left( x^2 + 4 \right)}\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{2x + 3}} \text{ dx }\]
\[\int\frac{x}{\left( x^2 + 2x + 2 \right) \sqrt{x + 1}} \text{ dx}\]
\[\int\frac{1}{\left( x + 1 \right) \sqrt{x^2 + x + 1}} \text{ dx }\]
If \[\int\frac{\cos 8x + 1}{\tan 2x - \cot 2x} dx\]
\[\int\frac{x^4 + x^2 - 1}{x^2 + 1} \text{ dx}\]
\[\int \text{cosec}^2 x \text{ cos}^2 \text{ 2x dx} \]
\[\int\frac{1}{\text{ cos }\left( x - a \right) \text{ cos }\left( x - b \right)} \text{ dx }\]
\[\int\frac{x^2}{\left( x - 1 \right)^3} dx\]
\[\int x\sqrt{1 + x - x^2}\text{ dx }\]
\[\int x^3 \left( \log x \right)^2\text{ dx }\]
