हिंदी

∫ a X 3 + B X X 4 + C 2 D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\frac{a x^3 + bx}{x^4 + c^2} dx\]
योग
Advertisements

उत्तर

\[\int\frac{\left( a x^3 + bx \right)}{x^4 + c^2}dx\]
\[ = \int\frac{a x^3}{x^4 + c^2}dx + \int\frac{bx}{\left( x^2 \right)^2 + c^2}dx\]
\[ = I_1 + I_2 \left(\text{ say } \right)\]
\[Where\]
\[ I_1 = \int \frac{a x^3}{x^4 + c^2}dx\ \text{and}\ I_2 = \int\frac{bx}{\left( x^2 \right)^2 + c^2}dx\]
\[Now, I_1 = \int\frac{a x^3}{x^4 + c^2}dx\]
\[\text{ let x }^4 + c^2 = t\]
\[ \Rightarrow 4 x^3 dx = dt\]
\[ \Rightarrow x^3 dx = \frac{dt}{4}\]
\[ I_1 = \frac{a}{4}\int\frac{dt}{t}\]
\[ = \frac{a}{4} \text{ log }\left| t \right| + C_1 \]
\[ = \frac{a}{4} \text{ log }\left| x^4 + c^2 \right| + C_1 \]
\[Now, I_2 = \int\frac{bx}{\left( x^2 \right)^2 + c^2}dx\]
\[\text{ let x} ^2 = p\]
\[ \Rightarrow \text{ 2x dx } = dp\]
\[ \Rightarrow \text { x dx }= \frac{dp}{2}\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.19 [पृष्ठ १०४]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.19 | Q 9 | पृष्ठ १०४

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\int\left\{ x^2 + e^{\log  x}+ \left( \frac{e}{2} \right)^x \right\} dx\]


\[\int\frac{x^5 + x^{- 2} + 2}{x^2} dx\]

\[\int \tan^2 \left( 2x - 3 \right) dx\]


\[\int\frac{1}{\text{cos}^2\text{ x }\left( 1 - \text{tan x} \right)^2} dx\]

\[\int\frac{x^2 + x + 5}{3x + 2} dx\]

\[\int\left( 5x + 3 \right) \sqrt{2x - 1} dx\]

\[\int\frac{- \sin x + 2 \cos x}{2 \sin x + \cos x} dx\]

\[\int\left( 2 x^2 + 3 \right) \sqrt{x + 2} \text{ dx  }\]

\[\int\frac{1}{\sqrt{a^2 - b^2 x^2}} dx\]

\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 + 1}} dx\]

` ∫  { x^2 dx}/{x^6 - a^6} dx `

\[\int\frac{1}{\sqrt{2x - x^2}} dx\]

\[\int\frac{\sin 8x}{\sqrt{9 + \sin^4 4x}} dx\]

\[\int\frac{\left( 3 \sin x - 2 \right) \cos x}{5 - \cos^2 x - 4 \sin x} dx\]

\[\int\sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]

`int 1/(sin x - sqrt3 cos x) dx`

\[\int x e^x \text{ dx }\]

\[\int e^\sqrt{x} \text{ dx }\]

\[\int\frac{\text{ log }\left( x + 2 \right)}{\left( x + 2 \right)^2}  \text{ dx }\]

\[\int\left( x + 1 \right) \text{ e}^x \text{ log } \left( x e^x \right) dx\]

\[\int e^x \cdot \frac{\sqrt{1 - x^2} \sin^{- 1} x + 1}{\sqrt{1 - x^2}} \text{ dx }\]

\[\int x^2 \sqrt{a^6 - x^6} \text{ dx}\]

\[\int x\sqrt{x^2 + x} \text{  dx }\]

\[\int\frac{1}{x\left( x^n + 1 \right)} dx\]

\[\int\frac{x}{\left( x - 3 \right) \sqrt{x + 1}} dx\]

The primitive of the function \[f\left( x \right) = \left( 1 - \frac{1}{x^2} \right) a^{x + \frac{1}{x}} , a > 0\text{ is}\]


\[\int\frac{x^4 + x^2 - 1}{x^2 + 1} \text{ dx}\]

\[\int\frac{1}{1 - x - 4 x^2}\text{  dx }\]

\[\int\frac{\sqrt{a} - \sqrt{x}}{1 - \sqrt{ax}}\text{  dx }\]

\[\int \tan^5 x\ \sec^3 x\ dx\]

\[\int\left( 2x + 3 \right) \sqrt{4 x^2 + 5x + 6} \text{ dx}\]

\[\int \sin^{- 1} \sqrt{x}\ dx\]

\[\int e^x \frac{\left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]

\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx}\]

\[\int\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} \text{ dx}\]

\[\int \left( e^x + 1 \right)^2 e^x dx\]


\[\int\frac{x^2 + 1}{x^2 - 5x + 6} \text{ dx }\]
 

\[\int\frac{x + 3}{\left( x + 4 \right)^2} e^x dx =\]


Find: `int (sin2x)/sqrt(9 - cos^4x) dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×