Advertisements
Advertisements
प्रश्न
उत्तर
\[\int\frac{\left( a x^3 + bx \right)}{x^4 + c^2}dx\]
\[ = \int\frac{a x^3}{x^4 + c^2}dx + \int\frac{bx}{\left( x^2 \right)^2 + c^2}dx\]
\[ = I_1 + I_2 \left(\text{ say } \right)\]
\[Where\]
\[ I_1 = \int \frac{a x^3}{x^4 + c^2}dx\ \text{and}\ I_2 = \int\frac{bx}{\left( x^2 \right)^2 + c^2}dx\]
\[Now, I_1 = \int\frac{a x^3}{x^4 + c^2}dx\]
\[\text{ let x }^4 + c^2 = t\]
\[ \Rightarrow 4 x^3 dx = dt\]
\[ \Rightarrow x^3 dx = \frac{dt}{4}\]
\[ I_1 = \frac{a}{4}\int\frac{dt}{t}\]
\[ = \frac{a}{4} \text{ log }\left| t \right| + C_1 \]
\[ = \frac{a}{4} \text{ log }\left| x^4 + c^2 \right| + C_1 \]
\[Now, I_2 = \int\frac{bx}{\left( x^2 \right)^2 + c^2}dx\]
\[\text{ let x} ^2 = p\]
\[ \Rightarrow \text{ 2x dx } = dp\]
\[ \Rightarrow \text { x dx }= \frac{dp}{2}\]
APPEARS IN
संबंधित प्रश्न
` ∫ sin x \sqrt (1-cos 2x) dx `
Integrate the following integrals:
If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]
\[\int\frac{1 - x^4}{1 - x} \text{ dx }\]
Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .