Advertisements
Advertisements
प्रश्न
\[\int\frac{1}{x^2 + 6x + 13} dx\]
योग
Advertisements
उत्तर
\[\int\frac{dx}{x^2 + 6x + 13}\]
\[ = \int\frac{dx}{x^2 + 2 \times x \times 3 + 9 - 9 + 13}\]
\[ = \int\frac{dx}{\left( x + 3 \right)^2 + 2^2}\]
\[ = \frac{1}{2} \tan^{- 1} \left( \frac{x + 3}{2} \right) + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{1}{\sqrt{1 - \cos 2x}} dx\]
\[\int\frac{1}{x (3 + \log x)} dx\]
` ∫ {sin 2x} /{a cos^2 x + b sin^2 x } ` dx
` ∫ {"cosec" x }/ { log tan x/2 ` dx
\[\int\frac{x^5}{\sqrt{1 + x^3}} dx\]
\[\ \int\ x \left( 1 - x \right)^{23} dx\]
` ∫ \sqrt{tan x} sec^4 x dx `
\[\int \sec^4 2x \text{ dx }\]
\[\int {cosec}^4 \text{ 3x } \text{ dx } \]
\[\int\frac{1}{\sin^4 x \cos^2 x} dx\]
` = ∫1/{sin^3 x cos^ 2x} dx`
\[\int\frac{x^2 - 1}{x^2 + 4} dx\]
\[\int\frac{1}{1 + x - x^2} \text{ dx }\]
\[\int\frac{x}{3 x^4 - 18 x^2 + 11} dx\]
\[\int\frac{1}{x\sqrt{4 - 9 \left( \log x \right)^2}} dx\]
\[\int\frac{\cos x}{\sqrt{\sin^2 x - 2 \sin x - 3}} dx\]
\[\int\frac{x^2}{x^2 + 6x + 12} \text{ dx }\]
\[\int\frac{6x - 5}{\sqrt{3 x^2 - 5x + 1}} \text{ dx }\]
\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]
\[\int x e^{2x} \text{ dx }\]
\[\int x^2 \text{ cos x dx }\]
\[\int \sec^{- 1} \sqrt{x}\ dx\]
\[\int \sin^{- 1} \sqrt{x} \text{ dx }\]
\[\int\frac{e^x \left( x - 4 \right)}{\left( x - 2 \right)^3} \text{ dx }\]
\[\int\frac{2x + 1}{\left( x + 1 \right) \left( x - 2 \right)} dx\]
\[\int\frac{x^2 + x - 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{1}{\cos x \left( 5 - 4 \sin x \right)} dx\]
\[\int\frac{x + 1}{x \left( 1 + x e^x \right)} dx\]
\[\int\frac{x^2 - 3x + 1}{x^4 + x^2 + 1} \text{ dx }\]
\[\int\frac{x^2 - 1}{x^4 + 1} \text{ dx }\]
\[\int\frac{x^2 + 1}{x^4 + 7 x^2 + 1} 2 \text{ dx }\]
\[\int\frac{1}{x^4 + 3 x^2 + 1} \text{ dx }\]
Write the anti-derivative of \[\left( 3\sqrt{x} + \frac{1}{\sqrt{x}} \right) .\]
\[\int\frac{1}{1 - \cos x - \sin x} dx =\]
\[\int \sec^2 x \cos^2 2x \text{ dx }\]
\[\int\frac{e^x - 1}{e^x + 1} \text{ dx}\]
\[\int\frac{\sin x}{\cos 2x} \text{ dx }\]
\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]
\[\int\frac{1 + x^2}{\sqrt{1 - x^2}} \text{ dx }\]
Find: `int (sin2x)/sqrt(9 - cos^4x) dx`
