∫ (√(x^3 + x^4 )) dx


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Question Number 77290 by TawaTawa last updated on 05/Jan/20

$\int \sqrt{x^{3} + x^{4}} dx$
Commented by mathmax by abdo last updated on 05/Jan/20

$l e t I = \int \sqrt{x^{3} + x^{4}} d x \Rightarrow I = \int \sqrt{x^{3} (1 + x)} d x = \int x^{\frac{3}{2}} \sqrt{1 + x} d x$ $c h a n g e m e n t \sqrt{1 + x} = t h i v e 1 + x = t^{2} \Rightarrow$ $I = \int {(t^{2} - 1)}^{\frac{3}{2}} t (2 t) d t = 2 \int t^{2} {(t^{2} - 1)}^{\frac{3}{2}} d t =_{t = c h (u)} 2 \int {c h}^{2} u {({s h}^{2} u)}^{\frac{3}{2}} s h (u) d u$ $= 2 \int {c h}^{2} u {s h}^{4} u d u = 2 \int {(\frac{s h (2 u)}{2})}^{2} {s h}^{2} u d u$ $= \frac{1}{2} \int {s h}^{2} (2 u) {s h}^{2} (u) d u = \frac{1}{2} \int \frac{c h (4 u) - 1}{2} \times \frac{c h (2 u) - 1}{2} d u$ $= \frac{1}{8} \int (c h (4 u) - 1) (c h (2 u) - 1) d u \Rightarrow$ $8 I = \int (c h (4 u) c h (2 u) - c h (4 u) - c h (2 u) + 1) d u$ $= \int \frac{(e^{4 u} + e^{- 4 u}) (e^{2 u} + e^{- 2 u})}{4} d u - \frac{1}{4} s h (4 u) - \frac{1}{2} s h (2 u) + u + c$ $= \frac{1}{4} \int (e^{6 u} + e^{2 u} + e^{- 2 u} + e^{- 6 u}) - \frac{1}{8} (e^{4 u} - e^{- 4 u}) - \frac{1}{2} (e^{2 u} - e^{- 2 u}) + u + c$ $= \frac{1}{24} e^{6 u} - \frac{1}{24} e^{- 6 u} + \frac{1}{8} e^{2 u} - \frac{1}{8} e^{- 2 u} - \frac{1}{8} e^{4 u} + \frac{1}{8} e^{- 4 u} - \frac{1}{2} e^{2 u}$ $+ \frac{1}{2} e^{- 2 u} + u + c$ $= \frac{1}{24} e^{6 l n (t + \sqrt{t^{2} - 1})} - \frac{1}{24} e^{- 6 l n (t + \sqrt{t^{2} - 1})} + \frac{1}{8} e^{2 l n (g + \sqrt{t^{2} - 1})} - \frac{1}{8} e^{- 2 l n (t + \sqrt{t^{2} - 1})}$ $- \frac{1}{8} e^{4 l n (t + \sqrt{t^{2} - 1})} + \frac{1}{8} e^{- 4 l n (t + \sqrt{t^{2} - 1})} - \frac{1}{2} e^{2 l n (t + \sqrt{t^{2} - 1})} + \frac{1}{2} e^{- 2 l n (t + \sqrt{t^{2} - 1})}$ $+ l n (t + \sqrt{t^{2} - 1}) + c$ $= \frac{1}{24} {(t + \sqrt{t^{2} - 1})}^{6} - \frac{1}{24 {(t + \sqrt{t^{2} - 1})}^{6}} + \frac{1}{8} {(t + \sqrt{t^{2} - 1})}^{2} - \frac{1}{8 {(t + \sqrt{t^{2} - 1})}^{2}}$ $- \frac{1}{8} {(t + \sqrt{t^{2} - 1})}^{4} + \frac{1}{8 {(t + \sqrt{t^{2} - 1})}^{4}} - \frac{1}{2} {(t + \sqrt{t^{2} - 1})}^{2} + \frac{1}{2 {(t + \sqrt{t^{2} - 1})}^{2}}$ $+ l n (t + \sqrt{t^{2} - 1}) + C$ $= \frac{1}{24} {(\sqrt{1 + x} + \sqrt{x})}^{6} - \frac{1}{24 {(\sqrt{1 + x} + \sqrt{x})}^{6}} + \frac{1}{8} {(\sqrt{1 + x} + \sqrt{x})}^{2}$ $- \frac{1}{8 {(\sqrt{1 + x} + \sqrt{x})}^{2}} - \frac{1}{8} {(\sqrt{1 + x} + \sqrt{x})}^{4} + \frac{1}{8 {(\sqrt{1 + x} + \sqrt{x})}^{4}} - \frac{1}{2} {(\sqrt{1 + x} + \sqrt{x})}^{2}$ $+ \frac{1}{2 {(\sqrt{1 + x} + \sqrt{x})}^{2}} + l n (\sqrt{1 + x} + \sqrt{x}) + C .$
Commented by TawaTawa last updated on 05/Jan/20

$God bless you sir .$
Commented by mathmax by abdo last updated on 05/Jan/20

$y o u a r e w e l c o m e .$
	Answered by petrochengula last updated on 05/Jan/20

	$= \int x \sqrt{x + x^{2}} d x$ $x + x^{2} = x^{2} + x + \frac{1}{4} - \frac{1}{4}$ $= {(x + \frac{1}{2})}^{2} - \frac{1}{4}$ $= \int x \sqrt{{(x + \frac{1}{2})}^{2} - \frac{1}{4}} d x$ $= \int x \sqrt{{(\frac{2 x + 1}{2})}^{2} - \frac{1}{4}} d x = \int x \sqrt{\frac{1}{4} ({(2 x + 1)}^{2} - 1)} d x$ $= \frac{1}{2} \int x \sqrt{{(2 x - 1)}^{2} - 1} d x$ $c o s h θ = 2 x - 1 \Rightarrow \frac{c o s h θ + 1}{2} = x$ $\frac{1}{2} s i n h θ d θ = d x$ $= \frac{1}{8} \int (c o s h θ + 1) {s i n h}^{2} θ d θ$ $= \frac{1}{8} \int {s i n h}^{2} θ c o s h θ d θ + \frac{1}{8} \int {s i n h}^{2} θ d θ$ $= \frac{1}{24} {s i n h}^{3} θ + \frac{1}{8} \int \frac{1}{2} (c o s h 2 θ - 1) d θ$ $= \frac{1}{24} {s i n h}^{3} θ + \frac{1}{16} \int (c o s h 2 θ - 1) d θ$ $= \frac{1}{24} {s i n h}^{3} θ + \frac{1}{32} s i n h 2 θ - \frac{1}{16} θ + C$ $= \frac{1}{24} {(s i n h ({c o s h}^{- 1} (2 x - 1)))}^{3} + \frac{1}{32} s i n h 2 ({c o s h}^{- 1} (2 x - 1)) - \frac{1}{16} {c o s h}^{- 1} (2 x - 1) + C$
	Commented by petrochengula last updated on 05/Jan/20

	$p l e a s e c h e c k$
	Commented by TawaTawa last updated on 05/Jan/20

	$God bless you sir$
	Commented by petrochengula last updated on 05/Jan/20

	$t h a n k s$
	Answered by MJS last updated on 05/Jan/20

	$\int \sqrt{x^{4} + x^{3}} d x = \int x \sqrt{x (x + 1)} d x =$ $[t = \sqrt{x (x + 1)} \to d x = \frac{2 \sqrt{x (x + 1)}}{2 x + 1} d t]$ $= \int t^{2} d t - \int \frac{t^{2}}{\sqrt{4 t^{2} + 1}} d t =$ $\int \frac{t^{2}}{\sqrt{4 t^{2} + 1}} d t =$ $[u = \sqrt{4 t^{2} + 1} \to d t = \frac{\sqrt{4 t^{2} + 1}}{4 t} d u]$ $= \frac{1}{8} \int \sqrt{u^{2} - 1} d u and this should be easy$ $= \frac{1}{3} t^{3} - \frac{1}{8} t \sqrt{4 t^{2} + 1} + \frac{1}{16} \ln (2 t + \sqrt{4 t^{2} + 1}) =$ $= \frac{1}{24} (8 x^{2} + 2 x - 3) \sqrt{x (x + 1)} + \frac{1}{16} \ln (2 x + 1 + 2 \sqrt{x (x + 1)}) + C$
	Commented by TawaTawa last updated on 05/Jan/20

	$God bless you sir$
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