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Question: Evaluate, \( y = \int {{e^{3a\log x}} + {e^{3x\log a}}dx} \)...

Evaluate, y=∫e3alog⁡x+e3xlog⁡adxy = \int {{e^{3a\log x}} + {e^{3x\log a}}dx}

Explanation

Solution

Here, first use the property of logarithm to simplify the terms given. Then using the formula of integration, find the result. In this question do not try to solve the integral directly, first simplify then integrate.

Complete step by step explanation:
y=∫(e3alog⁡x+e3xlog⁡a)dxy = \int {\left( {{e^{3a\log x}} + {e^{3x\log a}}} \right)dx}
Here, both terms are given in logarithm and exponential form, so first we have to simplify the terms using formula.
On separating the terms as two integrals
⇒y=∫e3alog⁡xdx+∫e3xlog⁡adx\Rightarrow y = \int {{e^{3a\log x}}dx + } \int {{e^{3x\log a}}dx}
Applying exponent property amn=(am)n{a^{mn}} = {\left( {{a^m}} \right)^n}
⇒y=∫(elog⁡x)3adx+∫(elog⁡a)3xdx\Rightarrow y = \int {{{\left( {{e^{\log x}}} \right)}^{3a}}dx + } \int {{{\left( {{e^{\log a}}} \right)}^{3x}}dx}
On simplifying
⇒y=∫(x)3adx+∫(a)3xdx\Rightarrow y = \int {{{\left( x \right)}^{3a}}dx + } \int {{{\left( a \right)}^{3x}}dx}

Now, we have simplified terms of function given, we can easily integrate using formula as it cannot be simplified more.

Using formula of integrals
y=x3a+13a+1+13a3xlog⁡a+Cy = \dfrac{{{x^{3a + 1}}}}{{3a + 1}} + \dfrac{1}{3}{a^{3x}}\log a + C

Note: In these types of questions, where logarithm and exponential both are given simplify the terms. We know that in this question all log function is to the base e, so it can be simplified directly using formula. Also we should use the formulae of exponents to simplify and arrange terms to make all the terms in simplest form. We know that exponential is the inverse of logarithm function; it means the logarithm of a given number x is the exponent to which another fixed number, the base b, must be raised, to produce that number x. In every type of question, first simplify then integrate, whether the function contains inverse trigonometric term, trigonometric term, exponential term, algebraic term etc. Simplification of function will make our integration less complicated and we will be able to find the result easily without any high level formulae of integration. One thing to keep in mind is that if we are given a logarithm function, then see whether the base is e or something else. If the base of log is different, then change its base to e.