Advertisements
Advertisements
प्रश्न
\[\int\left( \tan^{- 1} x^2 \right) x\ dx\]
बेरीज
उत्तर
Let I =
\[\int\] (tan–1 x2) x dx
Putting x2 = t
⇒ 2x dx = dt
Putting x2 = t
⇒ 2x dx = dt
\[\Rightarrow \text{ x dx }= \frac{dt}{2}\]
\[ \therefore I = \frac{1}{2}\int 1_{II} . \tan^{- 1_I} t . dt\]
\[ = \frac{1}{2} \tan^{- 1} t\int1 \text{ dt }- \int\left\{ \frac{d}{dt}\left( \tan^{- 1} t \right)\int1 dt \right\}dt\]
\[ = \frac{1}{2} \left[ \tan^{- 1} t . t - \int \frac{t}{1 + t^2}dt \right]\]
\[\text{ Now putting }\ 1 + t^2 = p\]
\[ \Rightarrow \text{ 2t dt }= dp\]
\[ \Rightarrow \text{ t dt} = \frac{dp}{2}\]
\[ \therefore I = \frac{1}{2}t . \tan^{- 1} t - \frac{1}{2}\int \frac{t dt}{1 + t^2}\]
\[ = \frac{t . \tan^{- 1} t}{2} - \frac{1}{2 x^2} \int \frac{dp}{p}\]
\[ = \frac{t . \tan^{- 1} t}{2} - \frac{1}{4}\ln p + C\]
\[ = \frac{x^2 . \tan^{- 1} x^2}{2} - \frac{1}{4} \text{ ln }\left| 1 + x^4 \right| + C \left[ \because p = 1 + t^2 \right]\]
\[ \therefore I = \frac{1}{2}\int 1_{II} . \tan^{- 1_I} t . dt\]
\[ = \frac{1}{2} \tan^{- 1} t\int1 \text{ dt }- \int\left\{ \frac{d}{dt}\left( \tan^{- 1} t \right)\int1 dt \right\}dt\]
\[ = \frac{1}{2} \left[ \tan^{- 1} t . t - \int \frac{t}{1 + t^2}dt \right]\]
\[\text{ Now putting }\ 1 + t^2 = p\]
\[ \Rightarrow \text{ 2t dt }= dp\]
\[ \Rightarrow \text{ t dt} = \frac{dp}{2}\]
\[ \therefore I = \frac{1}{2}t . \tan^{- 1} t - \frac{1}{2}\int \frac{t dt}{1 + t^2}\]
\[ = \frac{t . \tan^{- 1} t}{2} - \frac{1}{2 x^2} \int \frac{dp}{p}\]
\[ = \frac{t . \tan^{- 1} t}{2} - \frac{1}{4}\ln p + C\]
\[ = \frac{x^2 . \tan^{- 1} x^2}{2} - \frac{1}{4} \text{ ln }\left| 1 + x^4 \right| + C \left[ \because p = 1 + t^2 \right]\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\left\{ \sqrt{x}\left( a x^2 + bx + c \right) \right\} dx\]
\[\int\frac{x^{- 1/3} + \sqrt{x} + 2}{\sqrt[3]{x}} dx\]
\[\int\frac{x^5 + x^{- 2} + 2}{x^2} dx\]
\[\int\frac{1}{1 - \cos 2x} dx\]
\[\int\frac{1}{\sqrt{x + a} + \sqrt{x + b}} dx\]
\[\int\frac{3x + 5}{\sqrt{7x + 9}} dx\]
\[\int\frac{2 - 3x}{\sqrt{1 + 3x}} dx\]
\[\int\left( 5x + 3 \right) \sqrt{2x - 1} dx\]
` ∫ sin 4x cos 7x dx `
` ∫ {"cosec" x }/ { log tan x/2 ` dx
\[\int\sqrt{1 + e^x} . e^x dx\]
\[\int\frac{\cos^3 x}{\sqrt{\sin x}} dx\]
\[\int\frac{\left( x + 1 \right) e^x}{\sin^2 \left( \text{x e}^x \right)} dx\]
\[\int 5^{x + \tan^{- 1} x} . \left( \frac{x^2 + 2}{x^2 + 1} \right) dx\]
\[\int\frac{1}{\sin^3 x \cos^5 x} dx\]
Evaluate the following integrals:
\[\int\frac{x^7}{\left( a^2 - x^2 \right)^5}dx\]
\[\int\frac{x}{x^4 + 2 x^2 + 3} dx\]
\[\int\frac{x^2}{x^6 + a^6} dx\]
\[\int\frac{\sin 2x}{\sqrt{\sin^4 x + 4 \sin^2 x - 2}} dx\]
\[\int\frac{a x^3 + bx}{x^4 + c^2} dx\]
\[\int\frac{x^2 \left( x^4 + 4 \right)}{x^2 + 4} \text{ dx }\]
\[\int\frac{x^2}{x^2 + 6x + 12} \text{ dx }\]
\[\int\sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]
\[\int\frac{2x + 1}{\sqrt{x^2 + 4x + 3}} \text{ dx }\]
\[\int\frac{1}{\cos x \left( \sin x + 2 \cos x \right)} dx\]
\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]
\[\int\frac{1}{1 - \cot x} dx\]
\[\int x\ {cosec}^2 \text{ x }\ \text{ dx }\]
\[\int x^3 \tan^{- 1}\text{ x dx }\]
\[\int\frac{1}{x\left[ 6 \left( \log x \right)^2 + 7 \log x + 2 \right]} dx\]
\[\int\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{x^2 + 1}{x^4 + x^2 + 1} \text{ dx }\]
\[\int\frac{x^2 + 9}{x^4 + 81} \text{ dx }\]
\[\int\frac{\sin^6 x}{\cos^8 x} dx =\]
\[\int\frac{\sin^2 x}{\cos^4 x} dx =\]
\[\int\frac{x + 2}{\left( x + 1 \right)^3} \text{ dx }\]
\[\int \tan^5 x\ \sec^3 x\ dx\]
\[\int\frac{1 + x^2}{\sqrt{1 - x^2}} \text{ dx }\]
