86427
If \(a\lt 0\lt b\), then \(\int_{a}^{b} \frac{|x|}{x} d x\)
1 \(a-b\)
2 \(b-a\)
3 \(a+b\)
4 \(-a-b\)
Explanation:
(C) : Given, If \(\mathrm{a}\lt 0\lt \mathrm{b}\) then, \(\int_{a}^{b} \frac{|x|}{x} d x=\int_{a}^{0}\left[\frac{|x|}{x}\right] d x+\int_{0}^{b}\left[\frac{|x|}{x}\right] d x \quad=\int_{a}^{0}-1 d x+\int_{0}^{b} 1 . d x\) \(=[-\mathrm{x}]_{\mathrm{a}}^{0}+[\mathrm{x}]_{0}^{\mathrm{b}} .=(-1)(0-\mathrm{a})+\mathrm{b}=\mathrm{a}+\mathrm{b}\)
Rajasthan PET-2011
Integral Calculus
86378
\(\int \frac{\mathrm{e}^{\mathrm{x}^{2}}\left(2 \mathrm{x}+\mathrm{x}^{3}\right)}{\left(3+\mathrm{x}^{2}\right)^{2}} \mathrm{dx}\) is equal to :
86427
If \(a\lt 0\lt b\), then \(\int_{a}^{b} \frac{|x|}{x} d x\)
1 \(a-b\)
2 \(b-a\)
3 \(a+b\)
4 \(-a-b\)
Explanation:
(C) : Given, If \(\mathrm{a}\lt 0\lt \mathrm{b}\) then, \(\int_{a}^{b} \frac{|x|}{x} d x=\int_{a}^{0}\left[\frac{|x|}{x}\right] d x+\int_{0}^{b}\left[\frac{|x|}{x}\right] d x \quad=\int_{a}^{0}-1 d x+\int_{0}^{b} 1 . d x\) \(=[-\mathrm{x}]_{\mathrm{a}}^{0}+[\mathrm{x}]_{0}^{\mathrm{b}} .=(-1)(0-\mathrm{a})+\mathrm{b}=\mathrm{a}+\mathrm{b}\)
Rajasthan PET-2011
Integral Calculus
86378
\(\int \frac{\mathrm{e}^{\mathrm{x}^{2}}\left(2 \mathrm{x}+\mathrm{x}^{3}\right)}{\left(3+\mathrm{x}^{2}\right)^{2}} \mathrm{dx}\) is equal to :
86427
If \(a\lt 0\lt b\), then \(\int_{a}^{b} \frac{|x|}{x} d x\)
1 \(a-b\)
2 \(b-a\)
3 \(a+b\)
4 \(-a-b\)
Explanation:
(C) : Given, If \(\mathrm{a}\lt 0\lt \mathrm{b}\) then, \(\int_{a}^{b} \frac{|x|}{x} d x=\int_{a}^{0}\left[\frac{|x|}{x}\right] d x+\int_{0}^{b}\left[\frac{|x|}{x}\right] d x \quad=\int_{a}^{0}-1 d x+\int_{0}^{b} 1 . d x\) \(=[-\mathrm{x}]_{\mathrm{a}}^{0}+[\mathrm{x}]_{0}^{\mathrm{b}} .=(-1)(0-\mathrm{a})+\mathrm{b}=\mathrm{a}+\mathrm{b}\)
Rajasthan PET-2011
Integral Calculus
86378
\(\int \frac{\mathrm{e}^{\mathrm{x}^{2}}\left(2 \mathrm{x}+\mathrm{x}^{3}\right)}{\left(3+\mathrm{x}^{2}\right)^{2}} \mathrm{dx}\) is equal to :
86427
If \(a\lt 0\lt b\), then \(\int_{a}^{b} \frac{|x|}{x} d x\)
1 \(a-b\)
2 \(b-a\)
3 \(a+b\)
4 \(-a-b\)
Explanation:
(C) : Given, If \(\mathrm{a}\lt 0\lt \mathrm{b}\) then, \(\int_{a}^{b} \frac{|x|}{x} d x=\int_{a}^{0}\left[\frac{|x|}{x}\right] d x+\int_{0}^{b}\left[\frac{|x|}{x}\right] d x \quad=\int_{a}^{0}-1 d x+\int_{0}^{b} 1 . d x\) \(=[-\mathrm{x}]_{\mathrm{a}}^{0}+[\mathrm{x}]_{0}^{\mathrm{b}} .=(-1)(0-\mathrm{a})+\mathrm{b}=\mathrm{a}+\mathrm{b}\)
Rajasthan PET-2011
Integral Calculus
86378
\(\int \frac{\mathrm{e}^{\mathrm{x}^{2}}\left(2 \mathrm{x}+\mathrm{x}^{3}\right)}{\left(3+\mathrm{x}^{2}\right)^{2}} \mathrm{dx}\) is equal to :
