Exam 8: Techniques of Integration

The improper integral $\int _ { - \infty } ^ { \infty } \frac { 1 } { x ^ { 2 } + 1 } d x$ is

The integral $\int \frac { \sqrt { 4 + x } } { x } d x$ is

The improper integral $\int _ { 2 } ^ { \infty } \frac { 1 } { x \sqrt { x ^ { 2 } - 1 } } d x$ is

The integral $\int \frac { 1 } { \sqrt { 8 - 2 x - x ^ { 2 } } } d x$ is

The integral $\int x e ^ { 3 x } d x$ is

The number of subintervals needed to guarantee that the Trapezoidal Rule approximates $\int _ { 1 } ^ { 2 } \frac { e ^ { x } } { x } d x$ to within 0.0001 is

The integral $\int \frac { 1 } { \left( 4 x ^ { 2 } + 9 \right) ^ { \frac { 3 } { 2 } } } d x$ is

The integral $\int \tan ^ { 3 } x \sec x d x$ is

Using Simpson's Rule with n = 4, the approximated value of $\int _ { - 1 } ^ { 0 } \frac { 1 } { \sqrt { 1 - x ^ { 3 } } } d x,$ to three decimal places, is

The improper integral $\int _ { 0 } ^ { \infty } e ^ { - x } \sin x d x$ is

The integral $\int \frac { x ^ { 2 } } { 1 - x ^ { 4 } } d x$ is

The integral $\int ( \ln x ) ^ { 2 } d x$ is

Suppose you want to use the Midpoint Rule with n = 4 and equal-length subintervals to approximate $\int _ { 1 } ^ { 6 } e ^ { - x } d x$ ? What is the partition used for this approximation?

The integral $\int \frac { 1 } { \sqrt { x ^ { 2 } - 2 x - 3 } } d x$ is

The integral $\int \frac { 1 } { x ^ { 2 } - 6 x + 10 } d x$ is

The integral $\int \frac { 10 } { ( 1 - x ) \left( x ^ { 2 } + 4 \right) } d x$ is

The integral $\int \frac { x } { e ^ { x } } d x$ is

The integral $\int \sin ^ { - 1 } ( 2 x ) d x$ is

The integral $\int \frac { 1 } { \sqrt { 16 x ^ { 2 } + 1 } } d x$ is

The integral $\int \tan ^ { 3 } x \sec ^ { 2 } x d x$ is