Problem 12 Evaluate the following integrals... [FREE SOLUTION]

Chapter 1: Problem 12

Evaluate the following integrals: (i) \(\int \frac{x d x}{a+b x} d x\) (ii) \(\int \frac{2 x-1}{x-2} d x\) (iii) \(\int \frac{x^{2}}{1+x^{2}} d x\) (iv) \(\int \frac{x^{4}}{1+x^{2}} d x\)

Short Answer

Expert verified

Question: Evaluate the following integrals: (i) \(\int \frac{x}{a+bx} dx\) (ii) \(\int \frac{2x-1}{x-2} dx\) (iii) \(\int \frac{x^2}{1+x^2} dx\) (iv) \(\int \frac{x^4}{1+x^2} dx\) Answer: (i) \(\int \frac{x}{a+bx} dx = (a+bx) - a\ln|a + bx| + C\) (ii) \(\int \frac{2x-1}{x-2} dx = 2x - 3\ln|x-2| + C\) (iii) \(\int \frac{x^2}{1+x^2} dx = \frac{1}{2}(x^2 + 1 - \ln|x^2 + 1|) + C\) (iv) \(\int \frac{x^4}{1+x^2} dx = \frac{1}{3}x^3 - x + C\)

Step by step solution

- Step 1: Substitution

Let \(u = a + bx\). Then, \(du = b dx\), and we can solve for x and substitute it into the integrand.

- Step 2: Rewrite the integral

Rewrite the integral in terms of u: \(\int \frac{u-a}{ub} \frac{du}{b} = \int \frac{u-a}{u} du\)

- Step 3: Separate terms and integrate

Separate the terms and integrate: \(\int \frac{u-a}{u} du = \int (1 - \frac{a}{u}) du = u - a\ln|u| + C\)

- Step 4: Substitute back and simplify

Substitute back for u and simplify: \(\int \frac{x}{a+bx} dx = (a+bx) - a\ln|a + bx| + C\) #integral(ii)# (ii) \(\int \frac{2x-1}{x-2} dx\)

- Step 1: Polynomial long division

Perform polynomial long division: \(\frac{2x-1}{x-2} = 2 - \frac{3}{x - 2}\)

- Step 2: Separate terms and integrate

Separate the terms and integrate: \(\int (2 - \frac{3}{x-2}) dx = 2x - 3\ln|x-2| + C\) #integral(iii)# (iii) \(\int \frac{x^2}{1+x^2} dx\)

- Step 1: Substitution

Let \(u = x^2 + 1\). Then, \(du = 2x dx\), and we can substitute it into the integrand.

- Step 2: Rewrite the integral

Rewrite the integral in terms of u: \(\int \frac{x^2}{x^2 + 1} dx = \frac{1}{2} \int \frac{u-1}{u} du\)

- Step 3: Separate terms and integrate

Separate the terms and integrate: \(\frac{1}{2}\int \frac{u-1}{u} du = \frac{1}{2} \int (1 - \frac{1}{u}) du = \frac{1}{2}(u - \ln|u|) + C\)

- Step 4: Substitute back and simplify

Substitute back for u and simplify: \(\int \frac{x^2}{1+x^2} dx = \frac{1}{2}(x^2 + 1 - \ln|x^2 + 1|) + C\) #integral(iv)# (iv) \(\int \frac{x^4}{1+x^2} dx\)

- Step 1: Long division

Perform polynomial long division: \(\frac{x^4}{1+x^2} = x^2 - 1\)

- Step 2: Separate terms and integrate

Separate the terms and integrate: \(\int(x^2 - 1) dx = \frac{1}{3}x^3 - x + C\)

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91影视

Evaluate the following integrals: (i) \(\int \frac{x d x}{a+b x} d x\) (ii) \(\int \frac{2 x-1}{x-2} d x\) (iii) \(\int \frac{x^{2}}{1+x^{2}} d x\) (iv) \(\int \frac{x^{4}}{1+x^{2}} d x\)

Short Answer

Step by step solution

- Step 1: Substitution

- Step 2: Rewrite the integral

- Step 3: Separate terms and integrate

- Step 4: Substitute back and simplify

- Step 1: Polynomial long division

- Step 2: Separate terms and integrate

- Step 1: Substitution

- Step 2: Rewrite the integral

- Step 3: Separate terms and integrate

- Step 4: Substitute back and simplify

- Step 1: Long division

- Step 2: Separate terms and integrate

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