Advertisements
Advertisements
प्रश्न
\[\int \left( 3x + 4 \right)^2 dx\]
योग
Advertisements
उत्तर
\[\int \left( 3x + 4 \right)^2 dx\]
\[ = \int \left( 9 x^2 + 2 \times 3x \times 4 + 16 \right)dx\]
`= 9 ∫ x^2dx + 24 ∫ x dx + 16 ∫ dx`
\[ = 9\left[ \frac{x^3}{3} \right] + 24\left[ \frac{x^2}{2} \right] + 16x + C\]
\[ = 3 x^3 + 12 x^2 + 16x + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\left( x^e + e^x + e^e \right) dx\]
\[\int\sqrt{x}\left( 3 - 5x \right) dx\]
\[\int\frac{\left( x + 1 \right)\left( x - 2 \right)}{\sqrt{x}} dx\]
\[\int\frac{\cos x}{1 - \cos x} \text{dx }or \int\frac{\cot x}{\text{cosec } {x }- \cot x} dx\]
\[\int \left( a \tan x + b \cot x \right)^2 dx\]
\[\int\frac{1}{2 - 3x} + \frac{1}{\sqrt{3x - 2}} dx\]
\[\int\frac{x^2 + 5x + 2}{x + 2} dx\]
\[\int\frac{2x + 1}{\sqrt{3x + 2}} dx\]
` ∫ cos 3x cos 4x` dx
` ∫ {sec x "cosec " x}/{log ( tan x) }` dx
\[\int\frac{1}{\sqrt{a^2 + b^2 x^2}} dx\]
\[\int\frac{\cos x}{\sqrt{4 + \sin^2 x}} dx\]
\[\int\frac{\sin 2x}{\sqrt{\sin^4 x + 4 \sin^2 x - 2}} dx\]
\[\int\frac{\cos x}{\sqrt{4 - \sin^2 x}} dx\]
\[\int\frac{x + 7}{3 x^2 + 25x + 28}\text{ dx}\]
\[\int\frac{1}{4 \sin^2 x + 5 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{13 + 3 \cos x + 4 \sin x} dx\]
\[\int\frac{1}{3 + 4 \cot x} dx\]
\[\int x e^x \text{ dx }\]
\[\int x \cos^2 x\ dx\]
`int"x"^"n"."log" "x" "dx"`
\[\int \log_{10} x\ dx\]
\[\int\frac{x^3 \sin^{- 1} x^2}{\sqrt{1 - x^4}} \text{ dx }\]
\[\int\left( x + 2 \right) \sqrt{x^2 + x + 1} \text{ dx }\]
\[\int\frac{x^2 + x - 1}{x^2 + x - 6} dx\]
\[\int\frac{1}{x\left[ 6 \left( \log x \right)^2 + 7 \log x + 2 \right]} dx\]
\[\int\frac{3}{\left( 1 - x \right) \left( 1 + x^2 \right)} dx\]
\[\int\frac{1}{\sin x + \sin 2x} dx\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{x^2 + 1}} \text{ dx }\]
\[\int\frac{\cos 2x - 1}{\cos 2x + 1} dx =\]
If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then
\[\int \sin^4 2x\ dx\]
\[\int\frac{1}{\text{ sin} \left( x - a \right) \text{ sin } \left( x - b \right)} \text{ dx }\]
\[\int\frac{\sin x}{\cos 2x} \text{ dx }\]
\[\int\frac{1}{2 + \cos x} \text{ dx }\]
\[\int \left( x + 1 \right)^2 e^x \text{ dx }\]
\[\int\frac{x^2 + 1}{x^2 - 5x + 6} \text{ dx }\]
