Advertisements
Advertisements
प्रश्न
\[\int\limits_0^1 \frac{2x + 3}{5 x^2 + 1} dx\]
Advertisements
उत्तर
\[Let\ I = \int_0^1 \frac{2x + 3}{5 x^2 + 1}\ d x . Then, \]
\[I = \int_0^1 \frac{2x}{5 x^2 + 1} d x + \int_0^1 \frac{3}{5 x^2 + 1} d x\]
\[ \Rightarrow I = \frac{1}{5} \int_0^1 \frac{10x}{5 x^2 + 1} d x + 3 \int_0^1 \frac{1}{\left( \sqrt{5}x \right)^2 + 1^2} d x\]
\[ \Rightarrow I = \frac{1}{5} \left[ \log \left( 5 x^2 + 1 \right) \right]_0^1 + \frac{3}{\sqrt{5}} \left[ \tan^{- 1} \left( \sqrt{5}x \right) \right]_0^1 \]
\[ \Rightarrow I = \frac{1}{5} \log 6 + \frac{3}{\sqrt{5}} \tan^{- 1} \sqrt{5}\]
\[I = \int_0^1 \frac{2x}{5 x^2 + 1} d x + \int_0^1 \frac{3}{5 x^2 + 1} d x\]
\[ \Rightarrow I = \frac{1}{5} \int_0^1 \frac{10x}{5 x^2 + 1} d x + 3 \int_0^1 \frac{1}{\left( \sqrt{5}x \right)^2 + 1^2} d x\]
\[ \Rightarrow I = \frac{1}{5} \left[ \log \left( 5 x^2 + 1 \right) \right]_0^1 + \frac{3}{\sqrt{5}} \left[ \tan^{- 1} \left( \sqrt{5}x \right) \right]_0^1 \]
\[ \Rightarrow I = \frac{1}{5} \log 6 + \frac{3}{\sqrt{5}} \tan^{- 1} \sqrt{5}\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{\pi/2} \left( \sin x + \cos x \right) dx\]
\[\int\limits_0^{\pi/2} \cos^2 x\ dx\]
\[\int\limits_0^{\pi/2} x^2 \cos\ x\ dx\]
\[\int\limits_0^{\pi/4} x^2 \sin\ x\ dx\]
\[\int\limits_0^2 \frac{1}{\sqrt{3 + 2x - x^2}} dx\]
\[\int\limits_1^4 \frac{x^2 + x}{\sqrt{2x + 1}} dx\]
\[\int\limits_0^1 \frac{\sqrt{\tan^{- 1} x}}{1 + x^2} dx\]
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_0^{\pi/4} \frac{\tan^3 x}{1 + \cos 2x} dx\]
\[\int_0^\frac{\pi}{4} \frac{\sin x + \cos x}{3 + \sin2x}dx\]
\[\int\limits_0^{\pi/2} \cos^5 x\ dx\]
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_1^4 f\left( x \right) dx, where\ f\left( x \right) = \begin{cases}4x + 3 & , & \text{if }1 \leq x \leq 2 \\3x + 5 & , & \text{if }2 \leq x \leq 4\end{cases}\]
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
Evaluate the following integral:
\[\int\limits_{- 3}^3 \left| x + 1 \right| dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \sin x} dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^3 x\ dx\]
\[\int\limits_0^2 \left( x + 3 \right) dx\]
\[\int\limits_0^2 \left( x^2 + 1 \right) dx\]
\[\int\limits_1^2 x^2 dx\]
\[\int\limits_a^b \cos\ x\ dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^2 x\ dx .\]
\[\int\limits_0^1 \frac{1}{x^2 + 1} dx\]
\[\int\limits_{- 2}^1 \frac{\left| x \right|}{x} dx .\]
\[\int\limits_0^4 \frac{1}{\sqrt{16 - x^2}} dx .\]
\[\int\limits_2^3 \frac{1}{x}dx\]
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
\[\int\limits_0^\sqrt{2} \left[ x^2 \right] dx .\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^3 x} dx\] is equal to
\[\int\limits_0^1 \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\} dx\] is equal to
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^4 x dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
Which integral has lower and upper limits?
What is an antiderivative of \[2x+1\]?
