Let I=∫5x^2+4x+7(2x+3)^3/2dx Put 2x+3=t^2 Differentiating w.r.t x, we get 2dx=2t dt ∴ #160; dx=tdt From ( 1 ), we get x=t^2 ...

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Functions

∫5x2+4x+72x+3...

Question

$\int \frac{5 x^{2} + 4 x + 7}{(2 x + 3)^{\frac{3}{2}}}$

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Solution

Let $I = \int \frac{5 x^{2} + 4 x + 7}{(2 x + 3)^{\frac{3}{2}}} d x$
Put $2 x + 3 = t^{2}$
Differentiating w.r.t $x,$ we get
$2 d x = 2 t d t$
$∴$ $d x = t d t$
From ( 1 ), we get
$x = \frac{t^{2} - 3}{2}$
$∴$ $I = \int \frac{5 {(\frac{t^{2} - 3}{2})}^{2} + 4 (\frac{t^{2} - 3}{2}) + 7}{(t^{2})^{\frac{3}{2}}} . t d t$

$= \int \frac{5 (\frac{t^{4} - 6 t^{2} + 9}{4}) + 2 t^{2} - 6 + 7}{t^{3}} . t d t$

$= \int \frac{5 t^{4} - 30 t^{2} + 45 + 8 t^{2} + 4}{4 t^{3}} . t d t$

$= \int \frac{5 t^{4} - 22 t^{2} + 49}{4 t^{2}} d t$

$= \frac{5}{4} \int t^{2} d t - \frac{22}{4} \int d t + \frac{49}{4} \int t^{- 2} d t$

$= \frac{5}{4} . \frac{t^{3}}{3} - \frac{22}{4} t + \frac{49}{4} . \frac{t^{- 2 + 1}}{- 2 + 1} + c$

$= \frac{5}{12} t^{3} - \frac{22}{4} t - \frac{49}{4} . t^{- 1} + c$

$= \frac{5}{12} t^{3} - \frac{22}{4} t - \frac{49}{4 t} + c$

$∴$ $I = \frac{5}{12} (2 x + 3)^{\frac{3}{2}} - \frac{11}{2} \sqrt{2 x + 3} - \frac{49}{4} . \frac{1}{\sqrt{2 x + 3}} + c$

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