express each definite integral in terms of (u), but do not evaluate.\n1) (int_{-1}^{0}\frac{8x}{(4x^{2}+1)^{2…

express each definite integral in terms of (u), but do not evaluate.\n1) (int_{-1}^{0}\frac{8x}{(4x^{2}+1)^{2}}dx; u = 4x^{2}+1) 2) (int_{0}^{1}-12x^{2}(4x^{3}-1)^{3}dx; u = 4x^{3}-1)\n3) (int_{-1}^{2}6x(x^{2}-1)^{2}dx; u = x^{2}-1) 4) (int_{0}^{1}\frac{24x}{(4x^{2}+4)^{2}}dx; u = 4x^{2}+4)\nevaluate each definite integral.\n5) (int_{-3}^{0}-\frac{8x}{(2x^{2}+3)^{2}}dx; u = 2x^{2}+3) 6) (int_{0}^{1}\frac{16x}{(4x^{2}+4)^{2}}dx; u = 4x^{2}+4)\n7) (int_{-1}^{0}18x^{2}(3x^{3}+3)^{2}dx; u = 3x^{3}+3) 8) (int_{0}^{1}-\frac{8x}{(4x^{2}+2)^{2}}dx; u = 4x^{2}+2)

express each definite integral in terms of (u), but do not evaluate.\n1) (int_{-1}^{0}\frac{8x}{(4x^{2}+1)^{2}}dx; u = 4x^{2}+1) 2) (int_{0}^{1}-12x^{2}(4x^{3}-1)^{3}dx; u = 4x^{3}-1)\n3) (int_{-1}^{2}6x(x^{2}-1)^{2}dx; u = x^{2}-1) 4) (int_{0}^{1}\frac{24x}{(4x^{2}+4)^{2}}dx; u = 4x^{2}+4)\nevaluate each definite integral.\n5) (int_{-3}^{0}-\frac{8x}{(2x^{2}+3)^{2}}dx; u = 2x^{2}+3) 6) (int_{0}^{1}\frac{16x}{(4x^{2}+4)^{2}}dx; u = 4x^{2}+4)\n7) (int_{-1}^{0}18x^{2}(3x^{3}+3)^{2}dx; u = 3x^{3}+3) 8) (int_{0}^{1}-\frac{8x}{(4x^{2}+2)^{2}}dx; u = 4x^{2}+2)

Answer

Explanation:

Step1: Find the derivative of $u$

For $u = 4x^{2}+1$, then $du=8xdx$. When $x = - 1$, $u=4\times(-1)^{2}+1 = 5$; when $x = 0$, $u=4\times0^{2}+1 = 1$. So $\int_{-1}^{0}\frac{8x}{(4x^{2}+1)^{2}}dx=\int_{5}^{1}\frac{du}{u^{2}}$.

Step2: For $u = 4x^{3}-1$

$du = 12x^{2}dx$. When $x = 0$, $u=4\times0^{3}-1=-1$; when $x = 1$, $u=4\times1^{3}-1 = 3$. So $\int_{0}^{1}-12x^{2}(4x^{3}-1)^{3}dx=-\int_{-1}^{3}u^{3}du$.

Step3: For $u=x^{2}-1$

$du = 2xdx$, then $6xdx = 3du$. When $x=-1$, $u=(-1)^{2}-1 = 0$; when $x = 2$, $u=2^{2}-1=3$. So $\int_{-1}^{2}6x(x^{2}-1)^{2}dx=3\int_{0}^{3}u^{2}du$.

Step4: For $u = 4x^{2}+4$

$du=8xdx$, then $24xdx = 3du$. When $x = 0$, $u=4\times0^{2}+4 = 4$; when $x = 1$, $u=4\times1^{2}+4 = 8$. So $\int_{0}^{1}\frac{24x}{(4x^{2}+4)^{2}}dx=3\int_{4}^{8}\frac{du}{u^{2}}$.

Step5: For $u = 2x^{2}+3$

$du = 4xdx$, then $8xdx = 2du$. When $x=-3$, $u=2\times(-3)^{2}+3=21$; when $x = 0$, $u=2\times0^{2}+3 = 3$. So $\int_{-3}^{0}-\frac{8x}{(2x^{2}+3)^{2}}dx=-\int_{21}^{3}\frac{2du}{u^{2}}=\int_{3}^{21}\frac{2du}{u^{2}}$.

Step6: For $u = 4x^{2}+4$

$du=8xdx$, then $16xdx = 2du$. When $x = 0$, $u=4\times0^{2}+4 = 4$; when $x = 1$, $u=4\times1^{2}+4 = 8$. So $\int_{0}^{1}\frac{16x}{(4x^{2}+4)^{2}}dx=\int_{4}^{8}\frac{2du}{u^{2}}$.

Step7: For $u = 3x^{3}+3$

$du=9x^{2}dx$, then $18x^{2}dx = 2du$. When $x=-1$, $u=3\times(-1)^{3}+3 = 0$; when $x = 0$, $u=3\times0^{3}+3 = 3$. So $\int_{-1}^{0}18x^{2}(3x^{3}+3)^{2}dx=2\int_{0}^{3}u^{2}du$.

Step8: For $u = 4x^{2}+2$

$du = 8xdx$. When $x = 0$, $u=4\times0^{2}+2 = 2$; when $x = 1$, $u=4\times1^{2}+2 = 6$. So $\int_{0}^{1}-\frac{8x}{(4x^{2}+2)^{2}}dx=-\int_{2}^{6}\frac{du}{u^{2}}$.

  1. $\int_{5}^{1}\frac{du}{u^{2}}$
  2. $-\int_{-1}^{3}u^{3}du$
  3. $3\int_{0}^{3}u^{2}du$
  4. $3\int_{4}^{8}\frac{du}{u^{2}}$
  5. $\int_{3}^{21}\frac{2du}{u^{2}}$
  6. $\int_{4}^{8}\frac{2du}{u^{2}}$
  7. $2\int_{0}^{3}u^{2}du$
  8. $-\int_{2}^{6}\frac{du}{u^{2}}$