Search any question & find its solution
Question:
Answered & Verified by Expert
If $\int \frac{2 x+3}{x(x+1)(x+2)(x+3)+1} d x$ $=\frac{-1}{a x^2+b x+c}+\alpha$, then value of $a+b+c$ is equal to
Options:
Solution:
2005 Upvotes
Verified Answer
The correct answer is:
5
We have,
$$
\begin{aligned}
& \int \frac{2 x+3}{x(x+1)(x+2)(x+3)+1} d x=\frac{-1}{a x^2+b x+c}+\alpha \\
& \text { LHS }=\int \frac{2 x+3}{x(x+3)(x+1)(x+2)+1} d x \\
& \quad=\int \frac{(2 x+3) d x}{\left(x^2+3 x\right)\left(x^2+3 x+2\right)+1}
\end{aligned}
$$
Put $x^2+3 x=t \Rightarrow(2 x+3) d x=d t$
$$
\begin{aligned}
& \therefore \text { LHS }=\int \frac{d t}{t^2+2 t+1} \\
&=\int \frac{d t}{(t+1)^2}=-\frac{1}{t+1}+\alpha \\
&=\frac{-1}{x^2+3 x+1}+\alpha \\
& a=1, b=3, c=1 \\
& \therefore \quad a+b+c=1+3+1=5
\end{aligned}
$$
$$
\begin{aligned}
& \int \frac{2 x+3}{x(x+1)(x+2)(x+3)+1} d x=\frac{-1}{a x^2+b x+c}+\alpha \\
& \text { LHS }=\int \frac{2 x+3}{x(x+3)(x+1)(x+2)+1} d x \\
& \quad=\int \frac{(2 x+3) d x}{\left(x^2+3 x\right)\left(x^2+3 x+2\right)+1}
\end{aligned}
$$
Put $x^2+3 x=t \Rightarrow(2 x+3) d x=d t$
$$
\begin{aligned}
& \therefore \text { LHS }=\int \frac{d t}{t^2+2 t+1} \\
&=\int \frac{d t}{(t+1)^2}=-\frac{1}{t+1}+\alpha \\
&=\frac{-1}{x^2+3 x+1}+\alpha \\
& a=1, b=3, c=1 \\
& \therefore \quad a+b+c=1+3+1=5
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.