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Question Number 7792 by Tawakalitu. last updated on 15/Sep/16

Commented by FilupSmith last updated on 16/Sep/16
$(3) ∫_1 ^( 3) x(√(x^2 −1))dx =(1/2)∫_(x=1) ^( x=3) x(x^2 −1)^(1/2) dx u=x^2 −1 du=2xdx ∫_1 ^( 3) x(√(x^2 −1))dx=(1/2)∫_(x=1) ^( x=3) (√u)du =(1/2)∫_(x=1) ^( x=3) (u)^(1/2) du =(1/2)[(u)^(3/2) (2/3)]_(x=1) ^(x=3) =(1/3)[(x^2 −1)^(3/2) ]_(x=1) ^(x=3) =(1/3)[(9−1)^(3/2) −(1−1)^(3/2) ] =(1/3)[(9−1)^(3/2) −(1−1)^(3/2) ] =(1/3)(^( 2) (√8^3 ))=(1/3)×16(√2) =((16)/3)(√2) −−−−−−−−−−−−−−−−−−−−− (4) ∫ ((e^x −e^(−x) )/(e^x +e^(−x) )) e^x +e^(−x) =u du=e^x −e^x dx =∫(1/u)du =ln(e^x +e^(−x) )+c −−−−−−−−−−−−−−−−−−−−− (5) ∫_0 ^( 3) (1/(2x+1))dx=∫_0 ^( 3) (1/2)×(2/(2x+1))dx=(1/2)∫_0 ^( 3) (2/(2x+1))dx =(1/2)[ln(2x+1)]_0 ^3 =(1/2){ln(7)−ln(1)} =(1/2)ln(7) or ∫_0 ^( 3) (1/(2x+1))dx u=2x+1 du=2dx ⇒ dx=(1/2)du ∴∫_0 ^( 3) (1/(2x+1))dx=∫_0 ^( 3) (1/u)((1/2))du same as above$
$(3) \int_{1}^{3} x \sqrt{x^{2} - 1} d x$ $= \frac{1}{2} \int_{x = 1}^{x = 3} x {(x^{2} - 1)}^{\frac{1}{2}} d x$ $u = x^{2} - 1$ $d u = 2 x d x$ $\int_{1}^{3} x \sqrt{x^{2} - 1} d x = \frac{1}{2} \int_{x = 1}^{x = 3} \sqrt{u} d u$ $= \frac{1}{2} \int_{x = 1}^{x = 3} {(u)}^{\frac{1}{2}} d u$ $= \frac{1}{2} {[{(u)}^{\frac{3}{2}} \frac{2}{3}]}_{x = 1}^{x = 3}$ $= \frac{1}{3} {[{(x^{2} - 1)}^{\frac{3}{2}}]}_{x = 1}^{x = 3}$ $= \frac{1}{3} [{(9 - 1)}^{\frac{3}{2}} - {(1 - 1)}^{\frac{3}{2}}]$ $= \frac{1}{3} [{(9 - 1)}^{\frac{3}{2}} - {(1 - 1)}^{\frac{3}{2}}]$ $= \frac{1}{3} (^{2} \sqrt{8^{3}}) = \frac{1}{3} \times 16 \sqrt{2}$ $= \frac{16}{3} \sqrt{2}$ $- - - - - - - - - - - - - - - - - - - - -$ $(4) \int \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$ $e^{x} + e^{- x} = u$ $d u = e^{x} - e^{x} d x$ $= \int \frac{1}{u} d u$ $= \ln (e^{x} + e^{- x}) + c$ $- - - - - - - - - - - - - - - - - - - - -$ $(5) \int_{0}^{3} \frac{1}{2 x + 1} d x = \int_{0}^{3} \frac{1}{2} \times \frac{2}{2 x + 1} d x = \frac{1}{2} \int_{0}^{3} \frac{2}{2 x + 1} d x$ $= \frac{1}{2} {[\ln (2 x + 1)]}_{0}^{3}$ $= \frac{1}{2} {\ln (7) - \ln (1)}$ $= \frac{1}{2} \ln (7)$ $o r$ $\int_{0}^{3} \frac{1}{2 x + 1} d x$ $u = 2 x + 1$ $d u = 2 d x \Rightarrow d x = \frac{1}{2} d u$ $∴ \int_{0}^{3} \frac{1}{2 x + 1} d x = \int_{0}^{3} \frac{1}{u} (\frac{1}{2}) d u$ $s a m e a s a b o v e$
Commented by Tawakalitu. last updated on 16/Sep/16

$T h a n k y o u s i r . f o r y o u r h e l p .$
Commented by Tawakalitu. last updated on 16/Sep/16

$T h a n k s s o m u c h s i r, i r e a l l y a p p r e c i a t e .$
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