Answer
Verified
35.4k+ views
Hint: - Here we put into formula to solve easily$\left( {\int {\dfrac{1}{{{a^2} + {x^2}}}dx = \dfrac{1}{a}{{\tan }^{ - 1}}\dfrac{x}{a} + c} } \right)$.
$\int {\dfrac{1}{{{x^2} + 16}}dx} {\text{ }}$
We will use a substitution method to solve this problem, you may put a direct formula and get an answer.
Put $x{\text{ = 4tan}}\theta $
Differentiate with respect to$x$.
$dx = 4{\sec ^2}\theta d\theta $
Now,
$\int {\dfrac{{dx}}{{16 + {x^2}}}} = \int {\dfrac{{4{{\sec }^2}\theta d\theta }}{{16 + 16{{\tan }^2}\theta }} = \int {\dfrac{{4{{\sec }^2}\theta d\theta }}{{16\left( {1 + {{\tan }^2}\theta } \right)}}} } $
$\left( {\because 1 + {{\tan }^2}\theta = {{\sec }^2}\theta } \right)$
So, after cancel out, we get
$ \Rightarrow \dfrac{1}{4}\int {d\theta = \dfrac{1}{4}\theta + c = } \dfrac{1}{4}{\tan ^{ - 1}}\dfrac{x}{4} + c$ $\left( \begin{gathered}
\because x = 4\tan \theta \\
\therefore \theta = {\tan ^{ - 1}}\dfrac{x}{4} \\
\end{gathered} \right)$
Therefore the required answer is $\dfrac{1}{4}{\tan ^{ - 1}}\dfrac{x}{4} + c$
Note: - Whenever we face such type of integral question, we have to use substitution method to solve easily or this question can be solve by using direct formula$\left( {\int {\dfrac{1}{{{a^2} + {x^2}}}dx = \dfrac{1}{a}{{\tan }^{ - 1}}\dfrac{x}{a} + c} } \right)$.
$\int {\dfrac{1}{{{x^2} + 16}}dx} {\text{ }}$
We will use a substitution method to solve this problem, you may put a direct formula and get an answer.
Put $x{\text{ = 4tan}}\theta $
Differentiate with respect to$x$.
$dx = 4{\sec ^2}\theta d\theta $
Now,
$\int {\dfrac{{dx}}{{16 + {x^2}}}} = \int {\dfrac{{4{{\sec }^2}\theta d\theta }}{{16 + 16{{\tan }^2}\theta }} = \int {\dfrac{{4{{\sec }^2}\theta d\theta }}{{16\left( {1 + {{\tan }^2}\theta } \right)}}} } $
$\left( {\because 1 + {{\tan }^2}\theta = {{\sec }^2}\theta } \right)$
So, after cancel out, we get
$ \Rightarrow \dfrac{1}{4}\int {d\theta = \dfrac{1}{4}\theta + c = } \dfrac{1}{4}{\tan ^{ - 1}}\dfrac{x}{4} + c$ $\left( \begin{gathered}
\because x = 4\tan \theta \\
\therefore \theta = {\tan ^{ - 1}}\dfrac{x}{4} \\
\end{gathered} \right)$
Therefore the required answer is $\dfrac{1}{4}{\tan ^{ - 1}}\dfrac{x}{4} + c$
Note: - Whenever we face such type of integral question, we have to use substitution method to solve easily or this question can be solve by using direct formula$\left( {\int {\dfrac{1}{{{a^2} + {x^2}}}dx = \dfrac{1}{a}{{\tan }^{ - 1}}\dfrac{x}{a} + c} } \right)$.
Recently Updated Pages
To get a maximum current in an external resistance class 1 physics JEE_Main
f a body travels with constant acceleration which of class 1 physics JEE_Main
If the beams of electrons and protons move parallel class 1 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Let f be a twice differentiable such that fleft x rightfleft class 11 maths JEE_Main
Find the points of intersection of the tangents at class 11 maths JEE_Main
Other Pages
Oxidation state of S in H2S2O8 is A 6 B 7 C +8 D 0 class 12 chemistry JEE_Main
Two identical charged spheres suspended from a common class 12 physics JEE_Main
The mole fraction of the solute in a 1 molal aqueous class 11 chemistry JEE_Main
In a CO molecule the distance between C mass12 amu class 12 physics JEE_Main
A gas is compressed isothermally to half its initial class 11 physics JEE_Main
A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main