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Question: How do you integrate \[\int {{x^3}\sqrt {1 + {x^2}} } \] using integration by parts?...

How do you integrate ∫x31+x2\int {{x^3}\sqrt {1 + {x^2}} } using integration by parts?

Explanation

Solution

Hint : Here to integrate the given integral by using integral by parts defined as the ∫uv dx=u ∫v dx−∫u′  (∫v dx) dx\int {uv\,dx = u\,\int {v\,dx} - \int {u'\,\,\left( {\int {v\,dx} } \right)} } \,dx , where u is the function of u(x), v is the function of v(x) and u′u' is the derivative of the function u(x) , then simplify by using the standard integral and differentiation formulas.

Complete step-by-step answer :
Integration by parts is a method to find integrals of products, it is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways.
Integral parts can be defined as the integration of product of two function is equal to first function into integral of second function minus integral of derivative of first function into integral of second function i.e.,
∫uv dx=u ∫v dx−∫u′  (∫v dx) dx\int {uv\,dx = u\,\int {v\,dx} - \int {u'\,\,\left( {\int {v\,dx} } \right)} } \,dx
Here u is the first function, it is the function of u(x),
v is the second function, it is the function of v(x)
u′u' is the derivative of the function u(x)
Now, consider the given integral
⇒    ∫x31+x2  dx\Rightarrow \,\,\,\,\int {{x^3}\sqrt {1 + {x^2}} } \,\,dx
Here u=x3u = {x^3} , v=1+x2v = \sqrt {1 + {x^2}} and u′=dudx=ddxx3=3x2u' = \dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}{x^3} = 3{x^2}
Then , by the formula of integration by parts and it is indefinite integral then add constant term C known as integral constant.
⇒    ∫x31+x2  dx=x3∫1+x2 dx−∫ddxx3(∫1+x2 dx)dx+c\Rightarrow \,\,\,\,\int {{x^3}\sqrt {1 + {x^2}} } \,\,dx = {x^3}\int {\sqrt {1 + {x^2}} \,dx} - \int {\dfrac{d}{{dx}}{x^3}\left( {\int {\sqrt {1 + {x^2}} \,dx} } \right)dx} + c
⇒    ∫x31+x2  dx=x3∫(1+x2)12 dx−∫3x2(∫(1+x2)12dx)dx+c\Rightarrow \,\,\,\,\int {{x^3}\sqrt {1 + {x^2}} } \,\,dx = {x^3}\int {{{\left( {1 + {x^2}} \right)}^{\dfrac{1}{2}}}\,dx} - \int {3{x^2}\left( {\int {{{\left( {1 + {x^2}} \right)}^{\dfrac{1}{2}}}dx} } \right)dx} + c
As we know the formula ddxxn=nxn−1\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}} and ∫xndx=xn+1n+1+c\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c
⇒    ∫x31+x2  dx=x3(1+x2)12+1(12+1)−∫3x2((1+x2)12+1(12+1))dx+c\Rightarrow \,\,\,\,\int {{x^3}\sqrt {1 + {x^2}} } \,\,dx = {x^3}\dfrac{{{{\left( {1 + {x^2}} \right)}^{\dfrac{1}{2} + 1}}}}{{\left( {\dfrac{1}{2} + 1} \right)}} - \int {3{x^2}\left( {\dfrac{{{{\left( {1 + {x^2}} \right)}^{\dfrac{1}{2} + 1}}}}{{\left( {\dfrac{1}{2} + 1} \right)}}} \right)dx} + c
Therefore, 12+1=1+22=32\dfrac{1}{2} + 1 = \dfrac{{1 + 2}}{2} = \dfrac{3}{2}
⇒    ∫x31+x2  dx=x3(1+x2)32(32)−∫3x2( (1+x2)32(32))dx+c\Rightarrow \,\,\,\,\int {{x^3}\sqrt {1 + {x^2}} } \,\,dx = {x^3}\dfrac{{{{\left( {1 + {x^2}} \right)}^{\dfrac{3}{2}}}}}{{\left( {\dfrac{3}{2}} \right)}} - \int {3{x^2}\left( {\,\dfrac{{{{\left( {1 + {x^2}} \right)}^{\dfrac{3}{2}}}}}{{\left( {\dfrac{3}{2}} \right)}}} \right)dx} + c
⇒    ∫x31+x2  dx=x323(1+x2)32−∫3x223(1+x2)32dx+c\Rightarrow \,\,\,\,\int {{x^3}\sqrt {1 + {x^2}} } \,\,dx = {x^3}\dfrac{2}{3}{\left( {1 + {x^2}} \right)^{\dfrac{3}{2}}} - \int {3{x^2}\dfrac{2}{3}{{\left( {1 + {x^2}} \right)}^{\dfrac{3}{2}}}dx} + c
Simplifying and again, by applying the integration by parts to it.

⇒    ∫x31+x2  dx=x323(1+x2)32−∫3x223(1+x2)32dx+c ⇒    ∫x31+x2  dx=23x3(1+x2)32−∫2x2(1+x2)32dx+c ⇒∫x31+x2dx=23x3(1+x2)32−2x2(1+x2)5252+∫4x(1+x2)5252dx+c ⇒∫x31+x2dx=23x3(1+x2)32−2x2.25(1+x2)52+∫4x.25(1+x2)52dx+c ⇒∫x31+x2dx=23x3(1+x2)32−4x25(1+x2)52+∫8x5(1+x2)52dx+c   \Rightarrow \,\,\,\,\int {{x^3}\sqrt {1 + {x^2}} } \,\,dx = {x^3}\dfrac{2}{3}{\left( {1 + {x^2}} \right)^{\dfrac{3}{2}}} - \int {3{x^2}\dfrac{2}{3}{{\left( {1 + {x^2}} \right)}^{\dfrac{3}{2}}}dx} + c \\\ \Rightarrow \,\,\,\,\int {{x^3}\sqrt {1 + {x^2}} } \,\,dx = \dfrac{2}{3}{x^3}{\left( {1 + {x^2}} \right)^{\dfrac{3}{2}}} - \int {2{x^2}{{\left( {1 + {x^2}} \right)}^{\dfrac{3}{2}}}dx} + c \\\ \Rightarrow \int {{x^3}\sqrt {1 + {x^2}} dx = } \dfrac{2}{3}{x^3}{\left( {1 + {x^2}} \right)^{\dfrac{3}{2}}} - 2{x^2}\dfrac{{{{\left( {1 + {x^2}} \right)}^{\dfrac{5}{2}}}}}{{\dfrac{5}{2}}} + \int {4x\dfrac{{{{\left( {1 + {x^2}} \right)}^{\dfrac{5}{2}}}}}{{\dfrac{5}{2}}}dx + c} \\\ \Rightarrow \int {{x^3}\sqrt {1 + {x^2}} dx = } \dfrac{2}{3}{x^3}{\left( {1 + {x^2}} \right)^{\dfrac{3}{2}}} - 2{x^2}.\dfrac{2}{5}{\left( {1 + {x^2}} \right)^{\dfrac{5}{2}}} + \int {4x.\dfrac{2}{5}{{\left( {1 + {x^2}} \right)}^{\dfrac{5}{2}}}dx} + c \\\ \Rightarrow \int {{x^3}\sqrt {1 + {x^2}} dx = } \dfrac{2}{3}{x^3}{\left( {1 + {x^2}} \right)^{\dfrac{3}{2}}} - \dfrac{{4{x^2}}}{5}{\left( {1 + {x^2}} \right)^{\dfrac{5}{2}}} + \int {\dfrac{{8x}}{5}{{\left( {1 + {x^2}} \right)}^{\dfrac{5}{2}}}dx} + c \;

Again we apply the integration by parts we have

⇒∫x31+x2dx=23x3(1+x2)32−4x25(1+x2)52+8x5(1+x2)7272−∫85(1+x2)7272dx+c ⇒∫x31+x2dx=23x3(1+x2)32−4x25(1+x2)52+8x5.27(1+x2)72−∫85.27(1+x2)72dx+c ⇒∫x31+x2dx=23x3(1+x2)32−4x25(1+x2)52+16x35(1+x2)72−∫1635(1+x2)72dx+c   \Rightarrow \int {{x^3}\sqrt {1 + {x^2}} dx = } \dfrac{2}{3}{x^3}{\left( {1 + {x^2}} \right)^{\dfrac{3}{2}}} - \dfrac{{4{x^2}}}{5}{\left( {1 + {x^2}} \right)^{\dfrac{5}{2}}} + \dfrac{{8x}}{5}\dfrac{{{{\left( {1 + {x^2}} \right)}^{\dfrac{7}{2}}}}}{{\dfrac{7}{2}}} - \int {\dfrac{8}{5}\dfrac{{{{\left( {1 + {x^2}} \right)}^{\dfrac{7}{2}}}}}{{\dfrac{7}{2}}}dx} + c \\\ \Rightarrow \int {{x^3}\sqrt {1 + {x^2}} dx = } \dfrac{2}{3}{x^3}{\left( {1 + {x^2}} \right)^{\dfrac{3}{2}}} - \dfrac{{4{x^2}}}{5}{\left( {1 + {x^2}} \right)^{\dfrac{5}{2}}} + \dfrac{{8x}}{5}.\dfrac{2}{7}{\left( {1 + {x^2}} \right)^{\dfrac{7}{2}}} - \int {\dfrac{8}{5}.\dfrac{2}{7}{{\left( {1 + {x^2}} \right)}^{\dfrac{7}{2}}}dx} + c \\\ \Rightarrow \int {{x^3}\sqrt {1 + {x^2}} dx = } \dfrac{2}{3}{x^3}{\left( {1 + {x^2}} \right)^{\dfrac{3}{2}}} - \dfrac{{4{x^2}}}{5}{\left( {1 + {x^2}} \right)^{\dfrac{5}{2}}} + \dfrac{{16x}}{{35}}{\left( {1 + {x^2}} \right)^{\dfrac{7}{2}}} - \int {\dfrac{{16}}{{35}}{{\left( {1 + {x^2}} \right)}^{\dfrac{7}{2}}}dx} + c \;

Again we apply the integration by parts we have

⇒∫x31+x2dx=23x3(1+x2)32−4x25(1+x2)52+16x35(1+x2)72−1635(1+x2)9292+c ⇒∫x31+x2dx=23x3(1+x2)32−4x25(1+x2)52+16x35(1+x2)72−1635.29(1+x2)92+c ⇒∫x31+x2dx=23x3(1+x2)32−4x25(1+x2)52+16x35(1+x2)72−32315(1+x2)92+c   \Rightarrow \int {{x^3}\sqrt {1 + {x^2}} dx = } \dfrac{2}{3}{x^3}{\left( {1 + {x^2}} \right)^{\dfrac{3}{2}}} - \dfrac{{4{x^2}}}{5}{\left( {1 + {x^2}} \right)^{\dfrac{5}{2}}} + \dfrac{{16x}}{{35}}{\left( {1 + {x^2}} \right)^{\dfrac{7}{2}}} - \dfrac{{16}}{{35}}\dfrac{{{{\left( {1 + {x^2}} \right)}^{\dfrac{9}{2}}}}}{{\dfrac{9}{2}}} + c \\\ \Rightarrow \int {{x^3}\sqrt {1 + {x^2}} dx = } \dfrac{2}{3}{x^3}{\left( {1 + {x^2}} \right)^{\dfrac{3}{2}}} - \dfrac{{4{x^2}}}{5}{\left( {1 + {x^2}} \right)^{\dfrac{5}{2}}} + \dfrac{{16x}}{{35}}{\left( {1 + {x^2}} \right)^{\dfrac{7}{2}}} - \dfrac{{16}}{{35}}.\dfrac{2}{9}{\left( {1 + {x^2}} \right)^{\dfrac{9}{2}}} + c \\\ \Rightarrow \int {{x^3}\sqrt {1 + {x^2}} dx = } \dfrac{2}{3}{x^3}{\left( {1 + {x^2}} \right)^{\dfrac{3}{2}}} - \dfrac{{4{x^2}}}{5}{\left( {1 + {x^2}} \right)^{\dfrac{5}{2}}} + \dfrac{{16x}}{{35}}{\left( {1 + {x^2}} \right)^{\dfrac{7}{2}}} - \dfrac{{32}}{{315}}{\left( {1 + {x^2}} \right)^{\dfrac{9}{2}}} + c \;

Hence we have evaluated the integral.
So, the correct answer is “Option C”.

Note : The function can be integrated by using different methods. In integration we have two types: definite integral and indefinite integral. The definite integral has limit points and in the indefinite integral doesn’t have limit points. This problem can also be solved by using binomial expansion.