1)
∫ x 3 ( x 4 + 16 ) 3 d x \int x^3(x^4+16)^3dx ∫ x 3 ( x 4 + 16 ) 3 d x u = x 4 + 16 , d u = 4 x 3 d x , x 3 d x = 1 4 d u u=x^4+16, du=4x^3dx, x^3dx=\dfrac{1}{4}du u = x 4 + 16 , d u = 4 x 3 d x , x 3 d x = 4 1 d u
∫ x 3 ( x 4 + 16 ) 3 d x = 1 4 ∫ u 3 d u = 1 16 u 4 + C \int x^3(x^4+16)^3dx=\dfrac{1}{4}\int u^3du=\dfrac{1}{16}u^4+C ∫ x 3 ( x 4 + 16 ) 3 d x = 4 1 ∫ u 3 d u = 16 1 u 4 + C
= 1 16 ( x 4 + 16 ) 4 + C =\dfrac{1}{16}(x^4+16)^4+C = 16 1 ( x 4 + 16 ) 4 + C
2)
∫ 3 x 2 ( x 3 + 5 ) 7 d x \int 3x^2(x^3+5)^7dx ∫ 3 x 2 ( x 3 + 5 ) 7 d x u = x 3 + 5 , d u = 3 x 2 d x u=x^3+5, du=3x^2dx u = x 3 + 5 , d u = 3 x 2 d x
∫ 3 x 2 ( x 3 + 5 ) 7 d x = ∫ u 7 d u = 1 8 u 8 + C \int 3x^2(x^3+5)^7dx=\int u^7du=\dfrac{1}{8}u^8+C ∫ 3 x 2 ( x 3 + 5 ) 7 d x = ∫ u 7 d u = 8 1 u 8 + C
= 1 8 ( x 3 + 5 ) 8 + C =\dfrac{1}{8}(x^3+5)^8+C = 8 1 ( x 3 + 5 ) 8 + C
3)
∫ x 3 1 − x 4 d x \int \dfrac{x^3}{\sqrt{1-x^4}}dx ∫ 1 − x 4 x 3 d x u = 1 − x 4 , d u = − 4 x 3 d x , x 3 d x = − 1 4 d u u=1-x^4, du=-4x^3dx, x^3dx=-\dfrac{1}{4}du u = 1 − x 4 , d u = − 4 x 3 d x , x 3 d x = − 4 1 d u
∫ x 3 1 − x 4 d x = − 1 4 ∫ d u u = − 1 2 u + C \int \dfrac{x^3}{\sqrt{1-x^4}}dx=-\dfrac{1}{4}\int \dfrac{du}{\sqrt{u}}=-\dfrac{1}{2}\sqrt{u}+C ∫ 1 − x 4 x 3 d x = − 4 1 ∫ u d u = − 2 1 u + C
= − 1 2 1 − x 4 + C =-\dfrac{1}{2}\sqrt{1-x^4}+C = − 2 1 1 − x 4 + C
4)
∫ sin x ( cos x ) 3 d x \int\dfrac{\sin x}{(\cos x)^3}dx ∫ ( cos x ) 3 sin x d x u = cos x , d u = − sin x d x u=\cos x, du=-\sin xdx u = cos x , d u = − sin x d x
∫ sin x ( cos x ) 3 d x = − ∫ 1 u 3 d u = 1 2 u 2 + C \int\dfrac{\sin x}{(\cos x)^3}dx=-\int \dfrac{1}{u^3}du=\dfrac{1}{2u^2}+C ∫ ( cos x ) 3 sin x d x = − ∫ u 3 1 d u = 2 u 2 1 + C
= 1 2 ( cos x ) 2 + C =\dfrac{1}{2(\cos x)^2}+C = 2 ( cos x ) 2 1 + C
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