integrate x 3 x 1 4 dx Recall that 1 xa x a We can then rewrite our integral as dx x4 x 4dx Now recall that xadx where a 1 is equal to xa 1 a 1 C where C is the constant of integration Then x 4dx x 4 1 4 1 C 1 3x 3 C 1 3x3 C Answer link
1 Answer x 1 x4 dx 1 2 arctan x2 Method x 1 x4 dx 1 2 2xdx 1 x2 2 1 2 d x2 1 x2 2 I let x2 u I 1 2 du 1 u2 1 2arctanu 1 2arctan x2 Integrate using the power rule x3 x4 3 2dx 1 12 u3 C Reverse the substitution x3 x4 3 2dx 1 12 x4 3 3 C Please see the explanation Let u x 4 3 then du 4x 3dx or x 3dx 1 4 du intx 3 x 4 3 2dx 1 4intu 2du Integrate using the power rule intx 3 x 4 3 2dx 1 12 u 3 C Reverse the substitution intx 3
integrate x 3 x 1 4 dx
integrate x 3 x 1 4 dx
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This gives the following equations 1 3 B 0 B 1 3 C 1 3 1 C 2 3 This means that we can rewrite our original integral 1 1 x3 dx 1 3 1 x 1 x 2 x2 x 1 dx The first integral can be done using an explicit u substitution but it s rather clear that the answer is ln x 1 10317 views around the world You can reuse this answer
Explanation Substitute t 1 x2 dt 2xdx x3 1 x2dx 1 2 x2 1 x2 2xdx 1 2 t 1 tdt Now solve the integral in t 1 2 t 1 tdt 1 2 t tdt 1 2 tdt 1 2 t3 2dt 1 2 t1 2dt 1 2 t5 2 5 2 1 2 t3 2 3 2 C 1 5 t5 2 1 3 t3 2 C 1 2sqrt2 arc tan x 2 1 sqrt2 x 1 4sqrt2 ln x 2 sqrt2 x 1 x 2 sqrt2 x 1 C First let us consider the following Integrals I 1 int x 2 1 x 4 1 dx and
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Use the substitution x 2 4 t 2 Then 2x dx 2t dt So dx t x dt t t2 4 dt Now x3 x2 4dx t2 4 3 2 t2 t2 4 dt t2 4 t2dt t4 4t2 dt t5 5 t3 3 C t3 15 3t2 5 C 1 15 x2 4 3 2 3 x2 4 5 C 1 15 x2 4 3 2 3x2 7 C Answer link intx 3 sqrt 1 x 4 dx sqrt 1 x 4 2 C Using integration by substitution let u 1 x 4 du 4x 3dx Then together with the known formula intx ndx x n 1 n 1 C for n 1 we have intx 3 sqrt 1 x 4 dx 1 4int 4x 3 sqrt 1 x 4 dx 1 4int1 sqrt u du 1 4intu 1 2 du 1 4 u 1 2 1 2 C u 1 2 2 C sqrt 1 x 4 2 C
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