Question

Evaluate the following integrals: (i) \(\int \frac{5 x^{2}-12}{\left(x^{2}-6 x+13\right)^{2}} d x\) (ii) \(\int \frac{x^{3}+x-1}{\left(x^{2}+2\right)^{2}} d x\) (iii) \(\int \frac{x^{6}+x^{4}-4 x^{2}-2}{x^{3}\left(x^{2}+1\right)^{2}} d x\) (iv) \(\int \frac{d x}{x^{4}\left(x^{3}+1\right)^{2}}\)

Short answer

Question: Calculate the following integrals: (i) \(\int \frac{5 x^{2}-12}{\left(x^{2}-6 x+13\right)^{2}} d x\) Answer: \(\frac{1}{2}(10(\ln|x^2 -6x +13| - \frac{1}{x^2 - 6x + 13}(6x - 13)) - 12(-\frac{1}{x^2 - 6x + 13}) + C)\) (ii) \(\int \frac{x^{3}+x-1}{\left(x^{2}+2\right)^{2}} d x\) Answer: \((x^2 + 2)(\frac{-1}{2}(x^2 + 2 - 6)) - \int (x^2 - 1) \frac{1}{x^2 + 2} \cdot d(x^2 + 2) + C\) (iii) \(\int \frac{x^{6}+x^{4}-4 x^{2}-2}{x^{3}\left(x^{2}+1\right)^{2}} d x\) Answer: \(\int \frac{1}{x^3} dx + \int \frac{1}{(x^2 + 1)^2} dx + C\) (iv) \(\int \frac{d x}{x^{4}\left(x^{3}+1\right)^{2}}\) Answer: After applying the substitution method and finding the antiderivative, substitute back \(u = x^3\) and add the integration constant C. The integral cannot be simplified further without using techniques like partial fraction decomposition or integration by parts.

Step-by-step solution

Apply substitution method

Let \(u = x^2 - 6x + 13\). Then, \(du = (2x - 6)dx\). We can now rewrite the integral in terms of \(u\).

Rewrite the integral in terms of u

The integral becomes: \(\int \frac{(5x^2 - 12)dx}{(u)^{2}} = \frac{1}{2} \int \frac{(10x - 12)du}{(u)^{2}}\)

Break the integral into two simpler integrals

Now, we can split the integral into two simpler integrals: \(\frac{1}{2} \int \frac{10x - 12}{u^{2}}du = \frac{1}{2} \left(\int \frac{10x}{u^{2}}du - \int \frac{12}{u^{2}}du\right) = \frac{1}{2} \left(\int \frac{10(u + 6x -13)}{u^{2}}du - 12\int \frac{1}{u^{2}}du\right)\)

Integrate the two simpler integrals and substitute back x

Now, integrate both parts: \(\frac{1}{2} \left(10\int (1 - \frac{6x - 13}{u})du - 12\int u^{-2}du\right)\). We get \(\frac{1}{2} \left(10(\ln|u| - \frac{1}{u}(6x - 13) + C_1) - 12(-\frac{1}{u} + C_2)\right)\) as the antiderivative. Substituting back \(u = x^2 - 6x + 13\), we get \(\frac{1}{2}(10(\ln|x^2 -6x +13| - \frac{1}{x^2 - 6x + 13}(6x - 13)) - 12(-\frac{1}{x^2 - 6x + 13}) + C)\) as the final result. (ii) \(\int \frac{x^{3}+x-1}{\left(x^{2}+2\right)^{2}} d x\)

Apply substitution method with integration by parts

Let \(u = x^2 + 2\), \(du = 2xdx\). Then \(\int \frac{x^3 + x - 1}{(x^2+2)^2} dx = \int \frac{x( u - 2) + x - 1}{u^2} \cdot \frac{1}{2}du\)

Perform integration by parts

Integration by parts formula is: \(\int u dv = uv - \int v du\). In our case, we'll take \(u(x) = x(u-3)\) and \(dv(x) = \frac{1}{2}\frac{du}{u^2}\). Applying the integration by parts, we get: \(u(x)v(x) - \int v(x) du(x)\).

Calculate derivatives and antiderivatives, and substitute back x

Now, calculating the required derivatives and antiderivatives of \(u(x)\) and \(dv(x)\), we get \(u(x) = x(u-3)\); \(du(x) = (u-3)dx\) and \(v(x) =\frac{1}{2} \cdot -\frac{1}{u}\). Substitute \(u = x^2 + 2\), and integrate the remaining term: \(-\frac{1}{2} \int (u-3)\cdot 2\frac{dx}{u} = -\int (u-3) \frac{1}{u} \cdot d(x^2 + 2)\) The final result is given by \((x^2 + 2)(\frac{-1}{2}(x^2 + 2 - 6)) - \int (x^2 - 1) \frac{1}{x^2 + 2} \cdot d(x^2 + 2)+ C\). (iii) \(\int \frac{x^{6}+x^{4}-4 x^{2}-2}{x^{3}\left(x^{2}+1\right)^{2}} d x\)

Perform polynomial long division

To simplify the integrand, perform long division: \(\frac{x^6 + x^4 - 4x^2 - 2}{x^3(x^2 + 1)^2} = \frac{1}{x^3} + \frac{1}{(x^2 + 1)^2}\).

Integrate each part

Now, we will integrate each part separately: \(\int \frac{1}{x^3} dx + \int \frac{1}{(x^2 + 1)^2} dx\) The final result is given by \(\int \frac{1}{x^3} dx + \int \frac{1}{(x^2 + 1)^2} dx + C\). (iv) \(\int \frac{d x}{x^{4}\left(x^{3}+1\right)^{2}}\)

Apply substitution method

Let \(u = x^3\) and \(du = 3x^2 dx\), then the integral becomes: \(\int \frac{d(u^{1/3})}{(u^{4/3} + u^{1/3} + 1)^2} u^{-4} du\)

Rewrite the integral and calculate antiderivatives

The integral now becomes: \(\int \frac{1}{(u^{4/3}(u^{1/3} + 1))^2} du\). To find the antiderivative, we can use a standard table of integrals, or apply techniques like partial fraction decomposition and integration by parts. Then, substitute back \(u = x^3\), and add the integration constant to obtain the final result.