Advertisements
Advertisements
प्रश्न
\[\int\frac{1 - x^4}{1 - x} \text{ dx }\]
योग
उत्तर
\[\int\left( \frac{1 - x^4}{1 - x} \right)dx\]
\[ = \int\frac{\left( 1 - x^2 \right) \left( 1 + x^2 \right)}{\left( 1 - x \right)}dx\]
\[ = \int\frac{\left( 1 - x \right) \left( 1 + x \right) \left( 1 + x^2 \right)}{\left( 1 - x \right)}dx\]
\[ = \int\left( 1 + x \right) \left( 1 + x^2 \right)dx\]
\[ = \int\left( 1 + x^2 + x + x^3 \right)dx\]
\[ = x + \frac{x^3}{3} + \frac{x^2}{2} + \frac{x^4}{4} + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\left( 1 + x \right)^3}{\sqrt{x}} dx\]
\[\int\frac{1}{\sqrt{x}}\left( 1 + \frac{1}{x} \right) dx\]
\[\int\frac{x^6 + 1}{x^2 + 1} dx\]
\[\int\frac{x^5 + x^{- 2} + 2}{x^2} dx\]
\[\int \left( 3x + 4 \right)^2 dx\]
\[\int\frac{x^3 - 3 x^2 + 5x - 7 + x^2 a^x}{2 x^2} dx\]
\[\int\frac{\cos x}{1 + \cos x} dx\]
Integrate the following integrals:
\[\int\text{sin 2x sin 4x sin 6x dx} \]
\[\int\frac{a}{b + c e^x} dx\]
\[\int\frac{\cos 4x - \cos 2x}{\sin 4x - \sin 2x} dx\]
\[\int\frac{1 - \sin 2x}{x + \cos^2 x} dx\]
\[\int\frac{2 \cos 2x + \sec^2 x}{\sin 2x + \tan x - 5} dx\]
\[\int\frac{\sin 2x}{\sin 5x \sin 3x} dx\]
\[\int\frac{1}{1 + \sqrt{x}} dx\]
\[\int\frac{\cos^5 x}{\sin x} dx\]
\[\int\frac{x}{\sqrt{x^2 + a^2} + \sqrt{x^2 - a^2}} dx\]
\[\int \tan^3 \text{2x sec 2x dx}\]
\[\int\frac{1}{\sqrt{x} + x} \text{ dx }\]
\[\int\frac{2x - 1}{\left( x - 1 \right)^2} dx\]
\[\int\frac{1}{\sqrt{3 x^2 + 5x + 7}} dx\]
\[\int\frac{1}{\sqrt{\left( 1 - x^2 \right)\left\{ 9 + \left( \sin^{- 1} x \right)^2 \right\}}} dx\]
\[\int\frac{1}{1 - \cot x} dx\]
\[\int x \text{ sin 2x dx }\]
\[\int x \cos^2 x\ dx\]
\[\int\frac{x + \sin x}{1 + \cos x} \text{ dx }\]
\[\int x\left( \frac{\sec 2x - 1}{\sec 2x + 1} \right) dx\]
\[\int\left( \tan^{- 1} x^2 \right) x\ dx\]
\[\int e^x \left( \log x + \frac{1}{x^2} \right) dx\]
\[\int\frac{e^x}{x}\left\{ \text{ x }\left( \log x \right)^2 + 2 \log x \right\} dx\]
\[\int\sqrt{x^2 - 2x} \text{ dx}\]
\[\int\frac{x^2}{\left( x^2 + 1 \right) \left( 3 x^2 + 4 \right)} dx\]
\[\int\frac{\cos x}{\left( 1 - \sin x \right)^3 \left( 2 + \sin x \right)} dx\]
If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]
\[\int\frac{1}{1 + \tan x} dx =\]
\[\int\frac{\sin 2x}{a^2 + b^2 \sin^2 x}\]
\[\int\frac{\sin^6 x}{\cos x} \text{ dx }\]
\[\int\frac{1}{x \sqrt{1 + x^n}} \text{ dx}\]
\[\int\frac{5 x^4 + 12 x^3 + 7 x^2}{x^2 + x} dx\]
Find: `int (sin2x)/sqrt(9 - cos^4x) dx`
