$\int {\frac{{(2x + 1)}}{{{{({x^2}\, + 4x + 1)}^{\frac{3}{2}}}}}\,\,dx} $
- A
$\frac{{{x^3}}}{{{{({x^2}\, + 4x\, + 1)}^{\frac{1}{2}}}}}\,\, + \,\,C$
- ✓
$\frac{{{x}}}{{{{({x^2}\, + 4x\, + 1)}^{\frac{1}{2}}}}}\,\, + \,\,C$
- C
$\frac{{{x^2}}}{{{{({x^2}\, + 4x\, + 1)}^{\frac{1}{2}}}}}\,\, + \,\,C$
- D
$\frac{{{1}}}{{{{({x^2}\, + 4x\, + 1)}^{\frac{1}{2}}}}}\,\, + \,\,C$
✓
Answer
Correct option: B.$\frac{{{x}}}{{{{({x^2}\, + 4x\, + 1)}^{\frac{1}{2}}}}}\,\, + \,\,C$
b
$\int {\frac{{2x + 1}}{{{{({x^2} + 4x + 1)}^{\frac{3}{2}}}}}\,dx} $ $= \int {\frac{{2x + 1}}{{{x^3}\,{{\left( {1 + \,\frac{4}{x}\,\, + \,\frac{1}{{{x^2}}}} \right)}^{\frac{3}{2}}}}}\,dx} $ $=\int {\frac{{2{x^{ - 2}} + \,{x^{ - 3}}}}{{\,{{\left( {1 + \,\frac{4}{x}\,\, + \,\frac{1}{{{x^2}}}} \right)}^{\frac{3}{2}}}}}\,dx} $
now put $\frac{1}{{{x^2}}}\,\, + \,\,\frac{4}{x}\,\, + \,1\,\, = \,\,{t^2}\,$
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