Answer to Question #346625 in Calculus for Main

Question #346625

1. ∫ (π‘₯Β²+5)Β³ 2π‘₯𝑑π‘₯



2. ∫ 12π‘₯Β² √4π‘₯Β³ + 7𝑑π‘₯



3. ∫ √2π‘₯Β³+7 π‘₯²𝑑π‘₯



4. ∫ 5π‘₯√1 + 4π‘₯Β²



5. ∫(3π‘₯Β² βˆ’ 4π‘₯ + 2)Β² (3π‘₯ βˆ’ 2)𝑑π‘₯

1
Expert's answer
2022-06-01T02:59:48-0400

1.


"\\int(x^2+5)^32xdx"

"u=x^2+5, du=2xdx"

"\\int(x^2+5)^32xdx=\\int u^3du=\\dfrac{u^4}{4}+C""=\\dfrac{(x^2+5)^4}{4}+C"



2.


"\\int12x^2\\sqrt{4x^3+7}dx"

"u=4x^3+7, du=12x^2dx"



"\\int12x^2\\sqrt{4x^3+7}dx=\\int\\sqrt{u}du=\\dfrac{2}{3}u^{3\/2}+C""=\\dfrac{2}{3}(4x^3+7)^{3\/2}+C"



3.


"\\int(\\sqrt{2x^3+7})x^2dx"

"u=2x^3+7, du=6x^2dx"



"\\int(\\sqrt{2x^3+7})x^2dx=\\dfrac{1}{6}\\int\\sqrt{u}du=\\dfrac{1}{9}u^{3\/2}+C""=\\dfrac{1}{9}(2x^3+7)^{3\/2}+C"



4.


"\\int5x\\sqrt{1+4x^2}dx"

"u=1+4x^2, du=8xdx"



"\\int5x\\sqrt{1+4x^2}dx=\\dfrac{5}{8}\\int\\sqrt{u}du=\\dfrac{5}{12}u^{3\/2}+C""=\\dfrac{5}{12}(1+4x^2)^{3\/2}+C"



5.


"\\int(3x^2-4x+2)^2(3x-2)xdx"

"u=3x^2-4x+2, du=(6x-4)dx"

"\\int(3x^2-4x+2)^2(3x-2)xdx=\\dfrac{1}{2}\\int u^2du""=\\dfrac{u^3}{6}+C=\\dfrac{(3x^2-4x+2)^3}{6}+C"





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